How to solve $-20=15 \sin \theta- 30.98 \cos \theta$? 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.
 A: 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 $\theta$. 
A: If you're familiar with basic vectors, you may reason the $a\cos\theta + b\sin\theta $  formula like this
$$20 = 30.98 \cos \theta - 15\sin \theta =  \langle 30.98, -15 \rangle \bullet \langle \cos \theta, \sin\theta \rangle = C \cos(\theta - \alpha) $$
$C = \sqrt{30.98^2 + 15^2}$
$\alpha = \arctan\left(- \dfrac{30.98}{15}\right)$
