The general solution of `sinx - 3 sin 2x + sin3x = cos x - 3 cos 2x + cos 3x ` is.

Options

A. `npi+(pi)/(8)`
B. `(npi)/(2)+(pi)/(8)`
C. `(-1)^(n)(npi)/(2)+(pi)/(8)`
D. `2npi+cos^(-1)((3)/(2))`

Answer

B

Explanation

Given `sin 3x + sin x -3sin 2x = cos3x + cosx -3 cos 2x`
`rArr 2 sin2x cos x-3 sin 2x= 2 cos 2x cos x -3 cos 2x`
`rArr sin2x(2cos x-3) =cos 2x(2cos x-3)`
`rArr sin 2x(2cos x- 3) = cos2x (2cos x-3)" "[because 2 cos x - 3 ne 0]`
`rArr " "sin 2x = cos 2 x`
`rArr " " tan2x=1`
`rArr 2x = npi + (pi)/(4) rArr x = (npi)/(2) + (pi)/(8)`