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?

One easily verifies that
$$f(x):=\sin x+2\sin(2x)-\sin(3x)=2\sin x\bigl(2\cos x-\cos(2x)\bigr)\ .$$
Since the function $h(x):=2\cos x-\cos(2x)$ assumes its maximum ${3\over2}$ at $x={\pi\over3}$ (check this!) one obtains
$$f(x)\leq2\sin x \cdot {3\over2}=3\qquad(0<x<\pi)\ ,$$
whereby equality cannot take place, since $h\bigl({\pi\over2}\bigr)<{3\over2}$. 
The conclusion is that the given equation has no solutions in $\ ]0,\pi[\ $.
