The general solution of sin x - 3sin 2x + sin 3x = cos x -… — JEE Main 1989 Mathematics PYQ

1989

The general solution of

\sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x

A.
$n\pi + \dfrac{\pi}{8}$
B.
$\dfrac{n\pi}{2} + \dfrac{\pi}{8}$
C.
$(-1)^n \dfrac{n\pi}{2} + \dfrac{\pi}{8}$
D.
$2n\pi + \cos^{-1}\dfrac{3}{2}$

Show answer and solution

Correct option: B

Solution

Group as (sin3x + sinx) - 3sin2x = (cos3x + cosx) - 3cos2x.

Using sum-to-product: sin3x + sinx = 2sin2x·cosx, and cos3x + cosx = 2cos2x·cosx.

So: 2sin2x·cosx - 3sin2x = 2cos2x·cosx - 3cos2x
→ sin2x(2cosx - 3) = cos2x(2cosx - 3)
→ (2cosx - 3)(sin2x - cos2x) = 0

Since 2cosx - 3 = 0 has no real solution, we need sin2x = cos2x, i.e., tan2x = 1, so 2x = nπ + π/4, giving x = nπ/2 + π/8.

[1989]

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