Solving $2\sin(x+30^\circ) = \cos(x+150^\circ)$ for $x$ between $0^\circ$ and $360^\circ$ 
Solve this equation for $x$ between $0^\circ$ and $360^\circ$:
$$2\sin(x+30^\circ) = \cos(x+150^\circ)$$

Anyway thank you so much. This is my first question.

$$2\sin(x+30^{\circ}) = \cos(x+150^{\circ})$$
By the Cofunction Identity,
$$2\sin(x+30^{\circ}) = \sin(90^{\circ}-(x+150^{\circ}))$$
$$2\sin(x+30^{\circ}) = \sin(-x-60^{\circ})$$
$$2\sin(x+30^{\circ}) = \sin(-(x+60^{\circ}))$$
By the Negative Angle Identity,
$$2\sin(x+30^{\circ}) = -\sin(x+60^{\circ})$$
$$2\sin(x+30^{\circ})+\sin(x+60^{\circ}) = 0$$
By the Sum Identities,
$$2(\sin(x)\cos(30^{\circ})+\sin(30^{\circ})\cos(x))+\sin(x)\cos(60)+\sin(60)\cos(x)=0$$
$$\sqrt{3}\sin(x)+\cos(x)+\frac{1}{2}\sin(x)+\frac{\sqrt{3}}{2}\cos(x)=0$$
$$\sin(x)(\sqrt{3}+\frac{1}{2})=\cos(x)(-1-\frac{\sqrt{3}}{2})$$
$$\tan(x)=\frac{-1-\frac{\sqrt{3}}{2}}{\sqrt{3}+\frac{1}{2}}$$
$$\arctan{\frac{-1-\frac{\sqrt{3}}{2}}{\sqrt{3}+\frac{1}{2}}}=x$$
I'll let you use reference angles to calculate all the answers over the interval.
