First slide

Trigonometric equations

Question

The general solution of sin x - 3sin 2x + sin 3x = cos x -3cos 2x + cos 3x is

Moderate

A

$nπ + \frac{π}{8}$
B

$\frac{nπ}{2} + \frac{π}{8}$
C

$(- 1)^{n} \frac{nπ}{2} + \frac{π}{8}$
D

$2 nπ + \cos^{- 1} (\frac{3}{2})$

Solution

$\begin{array}{l} ∵ \sin x - 3 \sin 2 x + \sin 3 x = \cos x - 3 \cos 2 x + \cos 3 x \\ (\sin 3 x + \sin x) - 3 \sin 2 x = (\cos 3 x + \cos x) - 3 \cos 2 x \\ 2 \sin 2 x \cdot \cos x - 3 \sin 2 x = 2 \cos 2 x \cdot \cos x - 3 \cos 2 x \\ \sin 2 x (2 \cos x - 3) - \cos 2 x (2 \cos x - 3) = 0 \\ (\sin 2 x - \cos 2 x) (2 \cos x - 3) = 0 \\ So, \sin 2 x - \cos 2 x = 0 \\ as 2 \cos x - 3 \neq 0 \\ now, \sin 2 x = \cos 2 x \end{array}$
$\begin{array}{l} \tan 2 x = 1 = \tan \frac{π}{4} \\ 2 x = nπ + \frac{π}{4} \\ x = \frac{nπ}{2} + \frac{π}{8} \end{array}$

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