# Solve the Following Equation: Sin X − 3 Sin 2 X + Sin 3 X = Cos X − 3 Cos 2 X + Cos 3 X - Mathematics

Sum

Solve the following equation:
$\sin x - 3\sin2x + \sin3x = \cos x - 3\cos2x + \cos3x$

#### Solution

$\sin x - 3\sin2x + \sin3x = \cos x - 3\cos2x + \cos3x$
$\Rightarrow 2\sin2x\cos x - 3\sin2x = 2\cos2x\cos x - 3\cos2x$
$\Rightarrow \sin2x\left( 2\cos x - 3 \right) = \cos2x\left( 2\cos x - 3 \right)$
$\Rightarrow \left( \sin2x - \cos2x \right)\left( 2\cos x - 3 \right) = 0$
$\Rightarrow \sin2x - \cos2x = 0 or 2\cos x - 3 = 0$
$\Rightarrow \sin2x = \cos2x or \cos x = \frac{3}{2}$
$\Rightarrow \tan2x = 1 or \cos x = \frac{3}{2}$
But,
$\cos x = \frac{3}{2}$ is not possible.

$\left( - 1 \leq \cos x \leq 1 \right)$

$\therefore \tan2x = 1 = \tan\frac{\pi}{4}$

$\Rightarrow 2x = n\pi + \frac{\pi}{4}, n \in Z$

$\Rightarrow x = \frac{n\pi}{2} + \frac{\pi}{8}, n \in Z$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 11 Trigonometric equations
Exercise 11.1 | Q 7.5 | Page 22