[JEE Advanced 1989] the general solution of sin x 3 sin 2x sin 3x cos x 3
The general solution of $\sin x-3\sin2x+\sin3x=\cos x-3\cos2x+\cos3x$ is
(a) $n\pi+\frac\pi8$
(b) $\frac{n\pi}2+\frac\pi8$
(c) $(-1)^n\frac{n\pi}2+\frac\pi8$
(d) $2n\pi+\cos^{-1}\frac32$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
▲ 0
Correct answer: (b) $\frac{n\pi}2+\frac\pi8$
Group: $\sin x+\sin3x-3\sin2x=\sin2x(2\cos x-3)$ and similarly $\cos2x(2\cos x-3)$. Since $2\cos x-3\ne0$, $\sin2x=\cos2x\Rightarrow\tan2x=1\Rightarrow x=\frac{n\pi}2+\frac\pi8$.
JEE Advanced 1989 · Trigonometry — verified solution by the Vidaara Team.
Log in to post your own answer or join the discussion.
No comments yet — start the discussion.