Thank you in advance.
Good start. You did not apply the sum of angles formula for sine correctly. The sum of angles formula is:
$$\sin(\alpha+\beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$$
Applying that to $\sin(3x)$ gives:
$$\sin(3x) = \sin(2x)\cos x + \cos(2x)\sin x$$
Now, you can apply the double angle formula for $\sin (2x) = 2\sin x \cos x$ and for $\cos(2x) = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1-2\sin^2 x$.
Wolfram|Alpha: Computational Intelligence
According to Wolframalpha, there are no real solutions. Is it possible the problem statement has a typo? Maybe a plus should be a minus or vice versa?