Question

I am stuck on this homework problem.

All solutions in [0, 2π) for sin2+√3cosx=0.

That is how it is written, but is it correctly grouped as: sin(2+√3)cos(x)=0 ?

We are using double-angle identities.

Answer 1

Typo mistake. It must be sin2x.
\[\huge \sin2x+\sqrt{3}\cos x=0\]
\[\huge 2\sin x\cos x+\sqrt{3}\cos x=0\]
\[\huge \cos x(2\sin x+\sqrt{3})=0\]
\[\huge \cos x=0\]
OR
\[\huge \sin x=-\frac{\sqrt{3}}{2}\]
THEREFORE:
x=\frac{\pi}{2}, \frac{3\pi}{2}, \frac{4\pi}{3}, \frac{5\pi}{3}

Answer 2

\[\huge x=\frac{\pi}{2}, \frac{3\pi}{2}, \frac{4\pi}{3}, \frac{5\pi}{3}\]

Answer 3

Sorry about the last line in the first post. Rewrote it in my second post.

Answer 4

Thank you Azteck. The grouping confused me and I overlooked factoring the cos.