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?
On
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.
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[\ $.