Solving Different Types of Trigonometric Equations — Class 11 Trigonometry Notes

Introduction

While basic equations like $\sin x = k$ have standard solutions, many trigonometric equations require algebraic manipulation or the use of identities to reduce them to one of these basic forms. This section covers the primary techniques for solving more complex trigonometric equations.

1. Equations Reducible to Quadratic Form

Equations involving powers of a single trigonometric function (e.g., $2\sin^2x + 3\sin x + 1 = 0$ ) can be solved by making a substitution. By letting $y = \sin x$ , the equation becomes a standard quadratic equation ( $2y^2+3y+1=0$ ), which can be solved for 'y'. Once you find the values for 'y', substitute back ( $\sin x = y$ ) and solve for x, making sure to discard any solutions outside the range of the trigonometric function (e.g., $\sin x = 2$ is impossible).

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

This is a very common and important type of equation. The method to solve it is to convert the expression on the left into a single cosine or sine function.

Method:

Divide the entire equation by $\sqrt{a^2+b^2}$ .

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

The coefficients $\frac{a}{\sqrt{a^2+b^2}}$ and $\frac{b}{\sqrt{a^2+b^2}}$ can be represented as $\cos\alpha$ and $\sin\alpha$ for some angle $\alpha$ . The equation becomes $\cos\alpha\cos x + \sin\alpha\sin x = \frac{c}{\sqrt{a^2+b^2}}$ , which simplifies to $\cos(x-\alpha) = \frac{c}{\sqrt{a^2+b^2}}$ .
This is now a basic trigonometric equation that can be solved. For a solution to exist, we must have $|c| \le \sqrt{a^2+b^2}$ .

3. Equations Solved by Transformation

These equations are solved by using trigonometric identities to transform them into a simpler, solvable form.

Sum-to-Product: When an equation consists of sums or differences of sines or cosines (e.g., $\sin(5x)+\sin(x)=0$ ), use formulas like $\sin C + \sin D = 2\sin(\frac{C+D}{2})\cos(\frac{C-D}{2})$ . This converts the sum into a product, which is easy to solve by setting each factor to zero.
Product-to-Sum: Use formulas like $2\sin A\cos B = \sin(A+B)+\sin(A-B)$ to convert products into sums. This can sometimes help in simplifying or canceling terms.

4. Equations Solved by Squaring

Squaring both sides of an equation can be a useful technique, especially if it involves terms like $\sin x + \cos x$ or secant and tangent. However, this method can introduce extraneous roots (solutions that satisfy the squared equation but not the original one). It is absolutely essential to check all potential solutions by substituting them back into the original equation.

Example 1: Solving a Quadratic in $\tan x$

Question: Solve the equation $\tan^2x - (1+\sqrt{3})\tan x + \sqrt{3} = 0$ .

Solution: This is a quadratic equation in $\tan x$ .

Let $y = \tan x$ . The equation is $y^2 - (1+\sqrt{3})y + \sqrt{3} = 0$ .

This can be factored by inspection (sum of roots is $1+\sqrt{3}$ , product is $\sqrt{3}$ ):

$(y-1)(y-\sqrt{3})=0$ .

This gives two separate equations to solve:

Case 1: $\tan x = 1$ . The principal value is $\pi/4$ . The general solution is $x = n\pi + \pi/4$ .
Case 2: $\tan x = \sqrt{3}$ . The principal value is $\pi/3$ . The general solution is $x = n\pi + \pi/3$ .

The complete solution set is $x = n\pi + \pi/4$ or $x = n\pi + \pi/3$ , where $n \in \mathbb{Z}$ .

Example 2: Solving $a\cos x + b\sin x = c$

Question: Find the general solution of the equation $\sqrt{3}\cos x + \sin x = \sqrt{2}$ .

Solution: Here $a=\sqrt{3}, b=1$ . We divide the entire equation by $\sqrt{a^2+b^2} = \sqrt{(\sqrt{3})^2+1^2}=\sqrt{4}=2$ .

$\frac{\sqrt{3}}{2}\cos x + \frac{1}{2}\sin x = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

The LHS can be written in the form $\cos(A-B)$ . We can write it as $\cos(\pi/6)\cos x + \sin(\pi/6)\sin x$ , which is $\cos(x-\pi/6).$

The equation is now a basic one: $\cos(x-\pi/6) = \frac{1}{\sqrt{2}} = \cos(\pi/4)$ .

The general solution is $x-\pi/6 = 2n\pi \pm \pi/4$ .

This gives two sets of solutions:

Case 1: $x = 2n\pi + \pi/4 + \pi/6 = \mathbf{2n\pi + 5\pi/12}$ .
Case 2: $x = 2n\pi - \pi/4 + \pi/6 = \mathbf{2n\pi - \pi/12}$ .

Example 3: Using Sum-to-Product

Question: Solve $\sin(2x) + \sin(4x) + \sin(6x) = 0$ .

Solution: Group the first and third terms to make the subsequent factoring easier: $(\sin(6x)+\sin(2x)) + \sin(4x) = 0.$

Using the sum-to-product formula $\sin A + \sin B = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})$ $2\sin(\frac{6x+2x}{2})\cos(\frac{6x-2x}{2}) + \sin(4x) = 0$

$2\sin(4x)\cos(2x) + \sin(4x) = 0$ .

Factor out the common term $\sin(4x)$ : $\sin(4x)(2\cos(2x)+1)=0$ .

This gives two separate equations to solve:

$\sin(4x)=0 \implies 4x=n\pi \implies \mathbf{x = n\pi/4}$ .
$2\cos(2x)+1=0 \implies \cos(2x)=-1/2$ .

The principal value is $2\pi/3$ .

So, $2x = 2n\pi \pm 2\pi/3 \implies \mathbf{x=n\pi \pm \pi/3}$ .

Example 4: Solving by Squaring

Question: Solve $\sin x + \cos x = 1$ .

Solution: Squaring both sides: $(\sin x + \cos x)^2 = 1^2 \implies \sin^2x+\cos^2x+2\sin x\cos x = 1$ . Using identities, this becomes $1 + \sin(2x) = 1 \implies \sin(2x) = 0$ . The general solution for this is $2x = n\pi$ , so potential solutions are $x = n\pi/2$ . Now we must check these in the original equation for the interval $[0, 2\pi)$ to find a pattern.

$x=0$ : $\sin 0 + \cos 0 = 0+1=1$ (Correct).
$x=\pi/2$ : $\sin(\pi/2)+\cos(\pi/2)=1+0=1$ (Correct).
$x=\pi$ : $\sin(\pi)+\cos(\pi)=0+(-1)=-1$ (False - extraneous).
$x=3\pi/2$ : $\sin(3\pi/2)+\cos(3\pi/2)=-1+0=-1$ (False - extraneous). The valid solutions are of the form $x=2n\pi$ and $x=2n\pi+\pi/2$ .