How to solve $\sin(3x)+\cos(2x)={1\over2}$ I have a trouble to continuing this problem.
My work so far :
$$\sin(3x)+\cos(2x)={1\over2}$$
I try using this :
$$\sin(3x)+\sin(90^\circ-2x)={1\over2}$$
$$2\sin\left({3x+90^\circ-2x\over2}\right)\cos\left({3x-90^\circ+2x\over2}\right)={1\over2}$$
$$4\sin\left({x+90^\circ\over2}\right)\cos\left({5x-90^\circ\over2}\right)=1$$
How to solving this equation ?
Thanks for your help.
 A: Write $\sin x=s$
$$\dfrac12=3s-4s^3+1-2s^2\iff 8s^3+4s^2-6s-1=0$$
Use Cardano's formula
A: Recall that:
$$\sin(a+b) + \sin(a-b) = 2\sin(a)\cos(b).$$
Let $a=x$ and $b=2x$. Then:
$$\sin(x+2x) + \sin(x-2x) = 2\sin(x)\cos(2x) \Rightarrow\\
\sin(3x) + \sin(-x) = 2\sin(x)\cos(2x) \Rightarrow\\
\sin(3x) - \sin(x) = 2\sin(x)\cos(2x) \Rightarrow\\
\sin(3x) = 2\sin(x)\cos(2x) + \sin(x).$$
At this point, your equation can be rewritten as follows:
$$\sin(3x)+\cos(2x)={1\over2} \Rightarrow\\
2\sin(x)\cos(2x) + \sin(x) + \cos(2x) = {1\over2}.$$
Now, recall that $\cos(2x) = 1 - 2\sin^2(x).$ Then:
$$2\sin(x)(1-2\sin^2(x)) + \sin(x) + (1-2\sin^2(x)) = {1\over2} \Rightarrow \\
2\sin(x)-4\sin^3(x)+\sin(x)+1-2\sin^2(x)  = {1\over2} \Rightarrow \\
8\sin^3(x)+4\sin^2(x)-6\sin(x)-1 = 0.$$
To solve the last line, read the answer by lab bhattacharjee, which suggests you to use the Cardano's formula for finding the roots of third order polynomial, obtained with the substitution $s = \sin(x)$. Also, the Ruffini's rule is a good method for finding the roots of polynomials.
A: Uae that $$\sin(3x)=3\sin(x)-4\sin^3(x)$$
$$\cos(2x)=1-2\sin^2(x)$$
