Trigonometric Equations: Solving Methods

Introduction to Trigonometric Equations and Solutions

An equation that involves one or more trigonometric functions of an unknown angle is called a trigonometric equation. The unknown angle is typically represented by a variable such as $x$, $\theta$, $\phi$, etc.

Examples of trigonometric equations include:

• $ \sin x = \frac{1}{2} $

• $ \cos^2 \theta + \sin \theta = 1 $

• $ \tan 2y = \sqrt{3} $

• $ \sec^2 A - 2 \tan A = 0 $

Solving a trigonometric equation means finding the value(s) of the unknown angle that satisfy the equation.

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Solutions (or Roots) of Trigonometric Equations

A solution (or root) of a trigonometric equation is any value of the variable angle that makes the equation a true statement.

Let's consider the simple equation $ \sin x = \frac{1}{2} $.

From our knowledge of trigonometric ratios for specific angles, we know that $ \sin 30^\circ = \frac{1}{2} $. In radians, $ 30^\circ = \frac{\pi}{6} $. So, $ x = \frac{\pi}{6} $ is a solution.

Using the unit circle or our understanding of the sine function's graph, we also know that sine is positive in Quadrant II. The angle in Quadrant II with the same sine value as $ \frac{\pi}{6} $ is $ \pi - \frac{\pi}{6} = \frac{5\pi}{6} $. So, $ x = \frac{5\pi}{6} $ is another solution.

Since trigonometric functions are periodic, their values repeat at regular intervals. The sine function has a fundamental period of $ 2\pi $. This means that for any angle $ x $, $ \sin(x + 2n\pi) = \sin x $ for any integer $n$ ($n \in \mathbb{Z}$).

Therefore, if $ x = \frac{\pi}{6} $ is a solution, then any angle of the form $ \frac{\pi}{6} + 2n\pi $ is also a solution. For example, $ \frac{\pi}{6} + 2\pi = \frac{13\pi}{6} $, $ \frac{\pi}{6} - 2\pi = -\frac{11\pi}{6} $, $ \frac{\pi}{6} + 4\pi = \frac{25\pi}{6} $, etc., are all solutions.

Similarly, if $ x = \frac{5\pi}{6} $ is a solution, then any angle of the form $ \frac{5\pi}{6} + 2n\pi $ is also a solution. For example, $ \frac{5\pi}{6} + 2\pi = \frac{17\pi}{6} $, $ \frac{5\pi}{6} - 2\pi = -\frac{7\pi}{6} $, etc., are all solutions.

This periodic behaviour implies that most trigonometric equations, if they have any solutions, will have an infinite number of solutions.

Types of Solutions

To manage this infinite set of solutions, we classify them into two main types:

1. Principal Solutions: These are the solutions of a trigonometric equation that lie within a specific interval that covers exactly one period of the fundamental trigonometric functions. The standard interval used for principal solutions is $ [0, 2\pi) $ (or sometimes $ [0^\circ, 360^\circ) $). This interval corresponds to one full rotation on the unit circle starting from the positive x-axis. There are usually two principal solutions for most basic trigonometric equations within this interval, unless the solution lies on the boundary of the interval or is a value like 0, 1, or -1 for sine/cosine/tangent.

2. General Solution: This is a formula or expression that represents all possible solutions of a trigonometric equation. The general solution typically involves an arbitrary integer parameter, usually denoted by $n$ ($n \in \mathbb{Z}$), which accounts for the periodicity of the functions. For the equation $ \sin x = \frac{1}{2} $, the general solution combines all values of the form $ \frac{\pi}{6} + 2n\pi $ and $ \frac{5\pi}{6} + 2n\pi $.

When asked to "solve" a trigonometric equation, it generally means finding the general solution.

Example 1: Identifying Solutions

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Note for Competitive Exams

Principal Solutions of Trigonometric Equations

As defined in the previous section, the principal solutions of a trigonometric equation are those solutions that lie within the specific interval $ [0, 2\pi) $. This interval is chosen because it corresponds to one full cycle of the sine and cosine functions ($0^\circ$ to $360^\circ$, excluding $360^\circ$), ensuring that we find all unique solutions within a standard rotation before the values start repeating due to periodicity.

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Steps to Find Principal Solutions

To find the principal solutions of a trigonometric equation of the form $ \text{TrigFun}(x) = k $, where $\text{TrigFun}$ is one of the six trigonometric functions and $k$ is a constant, follow these steps:

1. Isolate the Trigonometric Function: Ensure the equation is in the form $ \sin x = k $, $ \cos x = k $, $ \tan x = k $, etc. This might require algebraic manipulation or using trigonometric identities.

2. Determine the Sign of $k$: Note whether the value $k$ is positive or negative.

3. Identify Possible Quadrants: Based on the sign of $k$ and the specific trigonometric function, determine the quadrant(s) where the angle $x$ could lie. Use the ASTC rule:

   • $ \sin x = k $: If $k > 0$, $x$ is in Quadrant I or II. If $k < 0$, $x$ is in Quadrant III or IV.

   • $ \cos x = k $: If $k > 0$, $x$ is in Quadrant I or IV. If $k < 0$, $x$ is in Quadrant II or III.

   • $ \tan x = k $: If $k > 0$, $x$ is in Quadrant I or III. If $k < 0$, $x$ is in Quadrant II or IV.

   • For reciprocal functions, use the same logic based on the sign of the reciprocal function.

4. Find the Reference Angle ($\alpha$): The reference angle is the acute angle ($ \alpha \in [0, \frac{\pi}{2}] $) formed by the terminal side of the angle $x$ and the x-axis. Find $\alpha$ such that $ \text{TrigFun}(\alpha) = |k| $. This is typically found using the inverse trigonometric function: $ \alpha = \arcsin|k| $, $ \alpha = \arccos|k| $, or $ \alpha = \arctan|k| $.

5. Calculate Solutions in $[0, 2\pi)$: Based on the identified quadrant(s) in Step 3 and the reference angle $\alpha$, determine the actual angle $x$ in the interval $ [0, 2\pi) $. The angle $x$ is related to $\alpha$ depending on the quadrant:

   • If $x$ is in Quadrant I, $ x = \alpha $.

   • If $x$ is in Quadrant II, $ x = \pi - \alpha $.

   • If $x$ is in Quadrant III, $ x = \pi + \alpha $.

   • If $x$ is in Quadrant IV, $ x = 2\pi - \alpha $.

   (Note: For angles on the axes, like $0, \pi/2, \pi, 3\pi/2$, these formulas might not apply directly, but their values are usually standard and easily identifiable, e.g., for $k=0, \pm 1$).

6. Verify: Check if the obtained values satisfy the original equation and lie within the specified interval $ [0, 2\pi) $.

Remember that for $ \sin x = k $ or $ \cos x = k $, solutions exist only if $ |k| \le 1 $. For $ \tan x = k $, solutions exist for any real value of $k$.

Example 1: Finding Principal Solutions (Cosine)

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Example 2: Finding Principal Solutions (Tangent)

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Note for Competitive Exams

General Solutions for $\sin x = k$, $\cos x = k$, $\tan x = k$

In the previous section, we discussed that trigonometric equations generally have infinitely many solutions due to the periodic nature of trigonometric functions. We also defined principal solutions as those within the interval $ [0, 2\pi) $. Now, we will learn how to find the general solution for the basic trigonometric equations of the form $\sin x = k$, $\cos x = k$, and $\tan x = k$.

A general solution is a formula, usually involving an arbitrary integer $n$ ($n \in \mathbb{Z} = \{\dots, -2, -1, 0, 1, 2, \dots\}$), that represents all possible values of the angle $x$ that satisfy the equation.

To find the general solution, we first find one particular solution of the equation. This particular solution is often taken from the principal value range of the corresponding inverse trigonometric function (e.g., $ [-\pi/2, \pi/2] $ for $\arcsin$, $ [0, \pi] $ for $\arccos$, $ (-\pi/2, \pi/2) $ for $\arctan$). Let's denote this particular solution by $\alpha$. So, $\sin \alpha = k$, $\cos \alpha = k$, or $\tan \alpha = k$. Then, we use the properties of the trigonometric functions and identities to find a general formula relating $x$ to $\alpha$.

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General Solution of $\sin x = \sin \alpha$

Consider the equation $\sin x = \sin \alpha$, where $\alpha$ is a known angle satisfying the equation $\sin \alpha = k$ (and $|k| \leq 1$).

Formula: The general solution for $\sin x = \sin \alpha$ is given by:

$ x = n\pi + (-1)^n \alpha $, where $ n \in \mathbb{Z} $

Derivation:

We are given $ \sin x = \sin \alpha $.

Rearrange the equation to set it equal to zero:

Use the Sum-to-Product identity for sine difference: $ \sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) $ (Equation 8.6). Here $C = x$ and $D = \alpha$.

$ \implies 2 \cos\left(\frac{x+\alpha}{2}\right) \sin\left(\frac{x-\alpha}{2}\right) = 0 $

For this product to be zero, at least one of the factors must be zero. This leads to two cases:

Case 1: $ \cos\left(\frac{x+\alpha}{2}\right) = 0 $

We know that $ \cos \theta = 0 $ when $ \theta $ is an odd multiple of $ \frac{\pi}{2} $. So, $ \theta = (2m+1)\frac{\pi}{2} $, where $ m $ is any integer ($ m \in \mathbb{Z} $).

Let $ \theta = \frac{x+\alpha}{2} $:

Multiply both sides by 2:

$ x+\alpha = (2m+1)\pi $

Solve for $x$:

Case 2: $ \sin\left(\frac{x-\alpha}{2}\right) = 0 $

We know that $ \sin \theta = 0 $ when $ \theta $ is an integer multiple of $ \pi $. So, $ \theta = m\pi $, where $ m \in \mathbb{Z} $.

Let $ \theta = \frac{x-\alpha}{2} $:

Multiply both sides by 2:

$ x-\alpha = 2m\pi $

Solve for $x$:

Now, we combine the solutions from Case 1 ($ x = (2m+1)\pi - \alpha $) and Case 2 ($ x = 2m\pi + \alpha $) into a single formula. Notice the pattern based on whether the multiple of $\pi$ is even or odd:

• When the multiple of $\pi$ is an even integer, say $2m$, the formula is $ x = 2m\pi + \alpha $.

• When the multiple of $\pi$ is an odd integer, say $2m+1$, the formula is $ x = (2m+1)\pi - \alpha $.

This pattern can be expressed using $ (-1)^n $. If $n$ is even, $ (-1)^n = 1 $, giving $ n\pi + \alpha $. If $n$ is odd, $ (-1)^n = -1 $, giving $ n\pi - \alpha $. Thus, we can write the combined solution as:

$ x = n\pi + (-1)^n \alpha $, where $ n \in \mathbb{Z} $.

This formula generates all possible solutions for $ \sin x = \sin \alpha $. When using this formula for an equation $\sin x = k$, $\alpha$ is any angle such that $\sin \alpha = k$, typically taken as the principal value $\arcsin k$ (which lies in $ [-\pi/2, \pi/2] $).

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General Solution of $\cos x = \cos \alpha$

Consider the equation $\cos x = \cos \alpha$, where $\alpha$ is a known angle satisfying the equation $\cos \alpha = k$ (and $|k| \leq 1$).

Formula: The general solution for $\cos x = \cos \alpha$ is given by:

$ x = 2n\pi \pm \alpha $, where $ n \in \mathbb{Z} $

Derivation:

We are given $ \cos x = \cos \alpha $.

Rearrange the equation:

Use the Sum-to-Product identity for cosine difference: $ \cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) $ (Equation 8.8). Here $C = x$ and $D = \alpha$.

$ \implies -2 \sin\left(\frac{x+\alpha}{2}\right) \sin\left(\frac{x-\alpha}{2}\right) = 0 $

This implies either $ \sin\left(\frac{x+\alpha}{2}\right) = 0 $ or $ \sin\left(\frac{x-\alpha}{2}\right) = 0 $ (since -2 is not zero).

Case 1: $ \sin\left(\frac{x+\alpha}{2}\right) = 0 $

We know that $ \sin \theta = 0 $ when $ \theta = m\pi $, where $ m \in \mathbb{Z} $.

So, $ \frac{x+\alpha}{2} = m\pi $.

Multiply by 2:

$ x+\alpha = 2m\pi $

Solve for $x$:

Case 2: $ \sin\left(\frac{x-\alpha}{2}\right) = 0 $

We know that $ \sin \theta = 0 $ when $ \theta = m\pi $, where $ m \in \mathbb{Z} $.

So, $ \frac{x-\alpha}{2} = m\pi $.

Multiply by 2:

$ x-\alpha = 2m\pi $

Solve for $x$:

Combining the results from Case 1 ($ x = 2m\pi - \alpha $) and Case 2 ($ x = 2m\pi + \alpha $):

These two forms represent solutions where $x$ differs from $\alpha$ by an even multiple of $\pi$ ($+ \alpha$ case) or differs from $-\alpha$ by an even multiple of $\pi$ ($-\alpha$ case). We can write this compactly as:

$ x = 2n\pi \pm \alpha $, where $ n \in \mathbb{Z} $.

This formula generates all possible solutions for $ \cos x = \cos \alpha $. When using this formula for an equation $\cos x = k$, $\alpha$ is any angle such that $\cos \alpha = k$, typically taken as the principal value $\arccos k$ (which lies in $ [0, \pi] $).

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General Solution of $\tan x = \tan \alpha$

Consider the equation $\tan x = \tan \alpha$, where $\alpha$ is a known angle satisfying the equation $\tan \alpha = k$ (and $k \in \mathbb{R}$).

Formula: The general solution for $\tan x = \tan \alpha$ is given by:

$ x = n\pi + \alpha $, where $ n \in \mathbb{Z} $

(Note: This formula is valid assuming $x \ne \frac{\pi}{2} + m\pi$ and $\alpha \ne \frac{\pi}{2} + m\pi$, where $m \in \mathbb{Z}$, as tangent is undefined at these values).

Derivation:

We are given $ \tan x = \tan \alpha $.

Using the quotient identity $ \tan \theta = \frac{\sin \theta}{\cos \theta} $, rewrite the equation:

Cross-multiply (assuming $ \cos x \ne 0 $ and $ \cos \alpha \ne 0 $):

$ \sin x \cos \alpha = \cos x \sin \alpha $

Rearrange to set the expression equal to zero:

Recognise the left side as the sine difference formula: $ \sin(A-B) = \sin A \cos B - \cos A \sin B $ (Equation 7.4). Here $A = x$ and $B = \alpha$.

We know that $ \sin \theta = 0 $ when $ \theta $ is an integer multiple of $ \pi $. So, $ \theta = n\pi $, where $ n \in \mathbb{Z} $.

Let $ \theta = x-\alpha $:

$ x-\alpha = n\pi $

Solve for $x$:

$ x = n\pi + \alpha $, where $ n \in \mathbb{Z} $

This formula generates all possible solutions for $ \tan x = \tan \alpha $. When using this formula for an equation $\tan x = k$, $\alpha$ is any angle such that $\tan \alpha = k$, typically taken as the principal value $\arctan k$ (which lies in $ (-\pi/2, \pi/2) $).

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Summary of General Solutions for Basic Equations

The general solutions for the three basic trigonometric equations are summarised below:

Equation	Condition on $k$	General Solution ($n \in \mathbb{Z}$)
$ \sin x = k $	$ |k| \leq 1 $	$ x = n\pi + (-1)^n \alpha $, where $ \sin \alpha = k $
$ \cos x = k $	$ |k| \leq 1 $	$ x = 2n\pi \pm \alpha $, where $ \cos \alpha = k $
$ \tan x = k $	$ k \in \mathbb{R} $	$ x = n\pi + \alpha $, where $ \tan \alpha = k $

In practice, for equations like $ \sin x = k $, $ \cos x = k $, or $ \tan x = k $, we first find a particular angle $\alpha$ (usually the principal value from inverse trigonometric functions) such that $ \sin \alpha = k $, $ \cos \alpha = k $, or $ \tan \alpha = k $. Then, we substitute this value of $\alpha$ into the respective general solution formula.

For reciprocal functions (cosec, sec, cot), convert the equation to its reciprocal form (sin, cos, tan) and then use the appropriate general solution formula.

For example, to solve $ \sec x = -2 $:

• Convert to cosine: $ \frac{1}{\cos x} = -2 \implies \cos x = -\frac{1}{2} $.

• Find a particular angle $\alpha$ such that $ \cos \alpha = -\frac{1}{2} $. We know that $ \cos \frac{2\pi}{3} = -\frac{1}{2} $. So, we can take $ \alpha = \frac{2\pi}{3} $.

• Use the general solution formula for $ \cos x = \cos \alpha $: $ x = 2n\pi \pm \alpha $.

• Substitute $\alpha$: The general solution is $ x = 2n\pi \pm \frac{2\pi}{3} $, where $ n \in \mathbb{Z} $.

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Examples

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Note for Competitive Exams

General Solutions for Other Trigonometric Equations

In the previous section, we found the general solutions for the basic trigonometric equations $ \sin x = k $, $ \cos x = k $, and $ \tan x = k $. We often encounter equations involving powers of trigonometric functions, particularly squares. The general solutions for equations like $ \sin^2 x = k $, $ \cos^2 x = k $, and $ \tan^2 x = k $ can be derived from the general solutions of the basic forms or using double angle identities.

Interestingly, the general solution form is the same for all three squared functions when equated to a squared value.

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General Solution of $\sin^2 x = \sin^2 \alpha$

Consider the equation $ \sin^2 x = \sin^2 \alpha $, where $\alpha$ is a known angle such that $ \sin^2 \alpha = k $ (and $0 \le k \le 1$).

Formula: The general solution for $ \sin^2 x = \sin^2 \alpha $ is given by:

$ x = n\pi \pm \alpha $, where $ n \in \mathbb{Z} $

Derivation:

We are given $ \sin^2 x = \sin^2 \alpha $.

Use the power reduction identity $ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} $ (derived from the cosine double angle formula, Equation 7.13):

Multiply both sides by 2:

$ 1 - \cos 2x = 1 - \cos 2\alpha $

Subtract 1 from both sides:

$ -\cos 2x = -\cos 2\alpha $

Multiply both sides by -1:

This equation is now in the basic form $ \cos \theta = \cos \beta $, where $ \theta = 2x $ and $ \beta = 2\alpha $. From the previous section, the general solution for $ \cos \theta = \cos \beta $ is $ \theta = 2n\pi \pm \beta $, where $ n \in \mathbb{Z} $.

Substitute back $ \theta = 2x $ and $ \beta = 2\alpha $:

Divide the entire equation by 2:

$ x = n\pi \pm \alpha $, where $ n \in \mathbb{Z} $

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General Solution of $\cos^2 x = \cos^2 \alpha$

Consider the equation $ \cos^2 x = \cos^2 \alpha $, where $\alpha$ is a known angle such that $ \cos^2 \alpha = k $ (and $0 \le k \le 1$).

Formula: The general solution for $ \cos^2 x = \cos^2 \alpha $ is given by:

$ x = n\pi \pm \alpha $, where $ n \in \mathbb{Z} $

Derivation:

We are given $ \cos^2 x = \cos^2 \alpha $.

Use the power reduction identity $ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} $ (derived from the cosine double angle formula, Equation 7.12):

Multiply both sides by 2:

$ 1 + \cos 2x = 1 + \cos 2\alpha $

Subtract 1 from both sides:

Again, this is in the form $ \cos \theta = \cos \beta $ where $ \theta = 2x $ and $ \beta = 2\alpha $. The general solution is $ \theta = 2n\pi \pm \beta $, where $ n \in \mathbb{Z} $.

Substitute back $ \theta = 2x $ and $ \beta = 2\alpha $:

Divide the entire equation by 2:

$ x = n\pi \pm \alpha $, where $ n \in \mathbb{Z} $

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General Solution of $\tan^2 x = \tan^2 \alpha$

Consider the equation $ \tan^2 x = \tan^2 \alpha $, where $\alpha$ is a known angle such that $ \tan^2 \alpha = k $ (and $k \ge 0$).

Formula: The general solution for $ \tan^2 x = \tan^2 \alpha $ is given by:

$ x = n\pi \pm \alpha $, where $ n \in \mathbb{Z} $

(Note: This assumes $x \ne \frac{\pi}{2} + m\pi$ and $\alpha \ne \frac{\pi}{2} + m\pi$, where $m \in \mathbb{Z}$, as tangent is undefined at these values).

Derivation (Method 1: Using Factoring)

We are given $ \tan^2 x = \tan^2 \alpha $.

Rearrange the equation:

$ \tan^2 x - \tan^2 \alpha = 0 $

Factor using the difference of squares formula $ a^2 - b^2 = (a-b)(a+b) $:

This product is zero if and only if one of the factors is zero:

Case 1: $ \tan x - \tan \alpha = 0 \implies \tan x = \tan \alpha $

From the previous section, the general solution for $ \tan x = \tan \alpha $ is $ x = m\pi + \alpha $, where $ m \in \mathbb{Z} $.

Case 2: $ \tan x + \tan \alpha = 0 \implies \tan x = -\tan \alpha $

Using the odd property of tangent, $ -\tan \alpha = \tan (-\alpha) $. So the equation becomes $ \tan x = \tan (-\alpha) $.

The general solution for this is $ x = k\pi + (-\alpha) = k\pi - \alpha $, where $ k \in \mathbb{Z} $.

Combining the solutions from Case 1 ($ x = m\pi + \alpha $) and Case 2 ($ x = k\pi - \alpha $):

The union of these two sets of solutions covers all possible values. We can express this combined solution using a single integer parameter $n$:

$ x = n\pi \pm \alpha $, where $ n \in \mathbb{Z} $

Derivation (Method 2: Using Double Angle Identity)

We are given $ \tan^2 x = \tan^2 \alpha $.

Use the identity $ \tan^2 \theta = \frac{1 - \cos 2\theta}{1 + \cos 2\theta} $. (This identity is derived from $\tan \theta = \sin \theta / \cos \theta$ and the double angle formulas for $\sin^2 \theta$ and $\cos^2 \theta$).

$ \implies \frac{1 - \cos 2x}{1 + \cos 2x} = \frac{1 - \cos 2\alpha}{1 + \cos 2\alpha} $ (assuming $ \cos 2x \ne -1 $ and $ \cos 2\alpha \ne -1 $)

Cross-multiply:

$ (1 - \cos 2x)(1 + \cos 2\alpha) = (1 - \cos 2\alpha)(1 + \cos 2x) $

Expand both sides:

$ 1 + \cos 2\alpha - \cos 2x - \cos 2x \cos 2\alpha = 1 + \cos 2x - \cos 2\alpha - \cos 2\alpha \cos 2x $

Subtract 1 from both sides and add $ \cos 2x \cos 2\alpha $ to both sides:

$ \cos 2\alpha - \cos 2x = \cos 2x - \cos 2\alpha $

Add $ \cos 2\alpha $ and $ \cos 2x $ to both sides to group terms:

$ 2 \cos 2\alpha = 2 \cos 2x $

Divide by 2:

As shown before, the general solution for $ \cos 2x = \cos 2\alpha $ is $ 2x = 2n\pi \pm 2\alpha $, which simplifies to:

$ x = n\pi \pm \alpha $, where $ n \in \mathbb{Z} $

Important Observation: The general solutions for $ \sin^2 x = \sin^2 \alpha $, $ \cos^2 x = \cos^2 \alpha $, and $ \tan^2 x = \tan^2 \alpha $ are all the same formula: $ x = n\pi \pm \alpha $, where $ n \in \mathbb{Z} $. This formula is very useful for solving equations involving squares of trigonometric functions.

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Examples

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Note for Competitive Exams

Solving Trigonometric Equations by Factoring and Using Identities

Not all trigonometric equations are as simple as $ \sin x = k $ or $ \cos x = k $. Many equations involve multiple trigonometric functions, functions of different angles (like $x$ and $2x$), or higher powers of trigonometric functions. To solve these more complex equations, we often need to employ algebraic techniques such as factoring and make strategic use of trigonometric identities to simplify them into forms that we already know how to solve.

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Solving by Factoring

If a trigonometric equation can be rearranged and factored into a product of expressions equal to zero, we can solve it by setting each factor equal to zero. This reduces a complex equation into a set of simpler, basic trigonometric equations.

Steps for Solving by Factoring:

1. Rearrange: Move all terms to one side of the equation so that the other side is zero.

2. Factor: Factor the trigonometric expression using common algebraic factoring techniques (e.g., common factors, difference of squares, factoring quadratics).

3. Set Factors to Zero: Set each factor containing a trigonometric function equal to zero.

4. Solve Simpler Equations: Solve each of the resulting basic trigonometric equations (like $ \sin x = k $, $ \cos x = k $, $ \tan x = k $). Find their general solutions.

5. Combine Solutions: The complete general solution is the union of the general solutions from each factor.

Caution: Avoid dividing both sides of the equation by a trigonometric expression unless you are certain that the expression can never be zero. If it can be zero, dividing by it will lead to losing potential solutions. Factoring is a safer approach.

Example 1: Solving a Quadratic Trigonometric Equation by Factoring

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Example 2: Solving by Factoring and Avoiding Division

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Solving using Trigonometric Identities

Many trigonometric equations cannot be solved by simple factoring alone. They might involve different angles or different trigonometric functions that cannot be directly factored. In such cases, we use trigonometric identities to rewrite the equation in a simpler form, ideally one that involves only a single trigonometric function and a single angle, or one that can be factored.

Common Uses of Identities in Solving Equations:

• Pythagorean Identities: To convert between $ \sin^2 x $ and $ \cos^2 x $. E.g., $ \sin^2 x = 1 - \cos^2 x $. This is useful for equations involving sums/differences of $ \sin^2 x $ and $ \cos^2 x $ or for creating quadratic forms.

• Double Angle Formulas: To relate functions of $2x$ to functions of $x$. E.g., $ \cos 2x $ can be replaced by $ \cos^2 x - \sin^2 x $, $ 2 \cos^2 x - 1 $, or $ 1 - 2 \sin^2 x $. The choice depends on other terms in the equation (e.g., if there is a $ \sin x $ term, use $ 1 - 2 \sin^2 x $ to get everything in terms of $ \sin x $).

• Sum and Difference Formulas: To expand or combine angles, e.g., $ \sin(x + \pi/4) $. Less common for initial simplification but useful in specific structures.

• Product-to-Sum / Sum-to-Product Formulas: To convert products into sums/differences or vice versa. Especially useful for equations involving sums of functions with different angles (as seen in the previous section) or products.

• Reciprocal and Quotient Identities: To express functions in terms of sine and cosine, e.g., replace $ \tan x $ with $ \sin x / \cos x $. This is often the first step if the equation involves multiple types of functions.

Example 3: Solving using a Double Angle Identity

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Equations of the Form $a \cos x + b \sin x = c$

Equations that are linear in $ \sin x $ and $ \cos x $ but cannot be easily factored can often be solved by transforming the expression $ a \cos x + b \sin x $ into a single trigonometric function, either $ R \cos(x - \alpha) $ or $ R \sin(x + \beta) $.

Method to solve $ a \cos x + b \sin x = c $ :

1. Calculate $ R = \sqrt{a^2 + b^2} $. This $R$ value will be the amplitude of the resulting single trigonometric function.

2. Divide the entire equation by $ R $:

   $ \frac{a}{\sqrt{a^2+b^2}} \cos x + \frac{b}{\sqrt{a^2+b^2}} \sin x = \frac{c}{\sqrt{a^2+b^2}} $

3. Find an angle $\alpha$ such that $ \cos \alpha = \frac{a}{\sqrt{a^2+b^2}} $ and $ \sin \alpha = \frac{b}{\sqrt{a^2+b^2}} $. Such an angle always exists because the sum of squares of coefficients is 1: $ (\frac{a}{R})^2 + (\frac{b}{R})^2 = \frac{a^2+b^2}{R^2} = \frac{a^2+b^2}{a^2+b^2} = 1 $. The angle $\alpha$ can be found using $ \tan \alpha = \frac{b}{a} $, making sure to select $\alpha$ in the correct quadrant based on the signs of $a$ and $b$.

4. Substitute these into the left side of the equation from Step 2:

   $ (\cos \alpha) \cos x + (\sin \alpha) \sin x = \frac{c}{R} $

5. Use the cosine difference identity $ \cos(A - B) = \cos A \cos B + \sin A \sin B $ (Equation 7.1) on the left side:

   $ \cos(x - \alpha) = \frac{c}{R} $

6. Solve this basic cosine equation: $ \cos(x - \alpha) = \frac{c}{\sqrt{a^2+b^2}} $. This equation has solutions only if $ |\frac{c}{\sqrt{a^2+b^2}}| \leq 1 $, i.e., $ c^2 \leq a^2 + b^2 $. If $|c| > \sqrt{a^2+b^2}$, there is no solution.

7. If solutions exist, let $\beta$ be a particular angle such that $ \cos \beta = \frac{c}{R} $ (e.g., $\beta = \arccos(c/R)$). The equation becomes $ \cos(x - \alpha) = \cos \beta $.

8. Using the general solution for cosine: $ x - \alpha = 2n\pi \pm \beta $, where $ n \in \mathbb{Z} $.

9. Solve for $x$: $ x = 2n\pi + \alpha \pm \beta $, where $ n \in \mathbb{Z} $.

Alternatively, we could let $ a = R \sin \beta $ and $ b = R \cos \beta $. Then $ \tan \beta = a/b $. The left side $ a \cos x + b \sin x $ becomes $ R (\sin \beta \cos x + \cos \beta \sin x) = R \sin(x + \beta) $. The equation becomes $ R \sin(x + \beta) = c $, or $ \sin(x + \beta) = c/R $. The general solution is $ x + \beta = n\pi + (-1)^n \gamma $, where $ \sin \gamma = c/R $. Then $ x = n\pi - \beta + (-1)^n \gamma $.

Example 4: Solving $a \cos x + b \sin x = c$

Answer:

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Note for Competitive Exams