<s>[INST] 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?
 [/INST] Here $$\sin x-\sin 3x+2\sin 2x = -2\cos(2x)\sin x+2\sin 2x\leq |2\cos 2x\sin x+2\sin 2x|$$
So $$\sin x-\sin 3x+2\sin 2x\leq 2|\sin 2x+\cos 2x|\leq 2\sqrt{2}|\sin (2x+45^\circ)|\leq 2\sqrt{2}<3$$
So no solution in $x\in (0,\pi)$
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