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

Question

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.

Answer 1

$$2\sin(x+30^{\circ}) = \cos(x+150^{\circ})$$ 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.

Answer 2

HINT

The equation is equivalent to

$$2\sin x \cos 30°+2\sin 30° \cos x=\cos x \cos 150°-\sin x\sin 150°$$

$$\sqrt 3 \sin x+\cos x=-\frac{\sqrt 3}2 \cos x-\frac12\sin x$$