Question

The general solution of sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x is ______.

पर्याय

• `nπ + π/8`

• `(nπ)/2 + π/8`

• `(-1)^n (nπ)/2 + π/8`

• `2nπ + cos^-1  3/2`

Answer 1

The general solution of sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x is `underlinebb((nπ)/2 + π/8)`.

Explanation:

We have, sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x

`\implies` (sin x + sin 3x) – 3 sin 2x = (cos x + cos 3x) – 3 cos 2x

`\implies` 2 sin 2x cos x – 3 sin 2x = 2 cos 2x cos x – 3 cos 2X

`[(∵ sin A + sin B = 2 sin ((A + B)/2) cos ((A - B)/2)),(and cos A + cos B = 2 cos ((A + B)/2) cos ((A - B)/2))]`

`\implies` (2 cos x – 3) sin 2x = (2 cos x – 3) cos 2x

`\implies` sin 2x = cos 2x

`\implies` tan 2x = 1

`\implies` tan 2x = `tan(nπ + π/4), n ∈ Z`

`\implies` 2x = `nπ + π/4`

`\implies` x = `(nπ)/2 + π/8`