TITLE: How to solve $-20=15 \sin \theta- 30.98 \cos \theta$?
QUESTION [5 upvotes]: How can I solve the following equation? $$-20=15 \sin \theta- 30.98 \cos \theta$$
I can't think of any way to solve it.  You can't factor out cosine because of the annoying little negative twenty, and if you divide by cosine you also get nowhere.

REPLY [6 votes]: There are many approaches. One way is to bring the $\sin\theta$ stuff to one side, and the rest to the other. Square both sides, and replace $\sin^2\theta$ by $1-\cos^2\theta$. We get a quadratic in $\cos\theta$. Solve, and for each solution check whether it is a solution of the original equation. 
For another approach, consider more generally $a\cos\theta+b\sin\theta=c$. Rewrite as
$$\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos\theta+\frac{b}{\sqrt{a^2+b^2}}\sin\theta\right)=c.$$ 
Find an angle $\varphi$ whose sine is $\frac{a}{\sqrt{a^2+b^2}}$ and whose cosine is 
$\frac{b}{\sqrt{a^2+b^2}}$.
Then our equation says that
$$\sin(\theta+\varphi)=\frac{c}{\sqrt{a^2+b^2}}.$$
Now find $\theta+\varphi$, and then $\theta$.