JEE Advanced 2014 · Paper 2, Mathematics question 46

Simplify. With $\sin3x=3\sin x-4\sin^3x$ and $\sin2x=2\sin x\cos x$:
$$\sin x+2\sin2x-\sin3x=2\sin x\left(2\sin^2x-1+2\cos x\right)=2\sin x\left(1+2\cos x-2\cos^2x\right).$$

Bound. $1+2c-2c^2$ has maximum $\frac32$ at $c=\frac12$, and $\sin x\le1$. So the left side is at most $2\times1\times\frac32=3$.

Equality would need $\sin x=1$ and $\cos x=\frac12$ at once, which is impossible. No solution.

(D)