Find the number of solutions of $\sin x+2 \sin 2x- \sin 3x=3$ In $(0 \:\:\pi)$Find the number of solutions of $$\sin x+2 \sin 2x- \sin 3x=3$$
The equation can be written as
$$\sin x+4 \sin x \cos x=3+\sin 3x$$ i.e.
$$\sin x(1+4\cos x)=3+\sin 3x$$ i.e.,
$$\sin x(1+4\cos x)=3+\sin x(3-4\sin^2 x)=3+\sin x(4\cos ^2x-1)$$  so
$$\sin x(4\cos^2 x-4\cos x-2)=-3$$ Any hint from here?

Hint of one possible solution: (assume your derivation is correct)
$$4\cos^2x-4\cos x-2=(2\cos x-1)^2-3\ge -3$$
and on $(0,\pi)$ we have $0<\sin x\le 1$.
Thus the LHS of your last equation should be $\ge -3$ on $(0,\pi)$.
Then you only need to check when the equality is satisfied, which is simple.
