$\sin(x/2) + \cos(x) = 0$, Solve for x if $0^\circ \leq x < 360^\circ$

Question

I have a pretty good grasp on what we're learning right now, but this question came up and I'm stumped. I would assume I need to use half-angle identities but would someone mind walking me through it please?

Thank you!

Answer 1

Try using double angle formulae.

$\cos(x)=\cos^2(x/2)-\sin^2(x/2)=1-2\sin^2(x/2)$

Therefore we have

$\sin(x/2)+\cos(x)=\sin(x/2)+1-2\sin^2(x/2)=0$

Let $u=\sin(x/2)$, so that you have a quadratic in $u$, which you can solve.

I'll leave you to do the rest.

Answer 2

We have $$ \cos+\sin\frac{x}{2}=0\iff \cos x=-\sin\frac{x}{2}=\cos\frac{x+\pi}{2}. $$ It follows that $$ x=\frac{x}{2}+\frac\pi2+2k\pi \, \text{ or }\, x=-\frac{x}{2}-\frac\pi2+2k\pi\quad k \in \mathbb{Z}, $$ i.e. $$ x=\pi+4k\pi\, \text{ or }\, x=-\frac\pi3+\frac43k\pi \quad k \in \mathbb{Z} $$