# Solving Trigonometric Equations with a ranging from $-2\pi$ to $2\pi$

Solve the following for $-2\pi \le \theta \le 2\pi$: $$12\cos(2\theta) – 6 = \sin\theta$$

What I did was used trig identities to make cosine into sine by using $\cos^2\theta = 1 - sin^2(\theta)$ and solving from there. I ended up getting $\theta = -49°$ and $\theta=42°$. However, when it comes to the range, when you go into the negatives, does the CAST rule become reversed? If you could please help me solve this, it would be greatly appreciated!

$12\cos^2\theta-12\sin^2\theta-6=\sin\theta \iff -24\sin^2\theta+6=\sin\theta$.
Now you have a quadratic equation over $\sin$. You solve it, get the values of $\sin$ that satisfy it, and then get the values of $\theta$.
you need the double angle formula $$\cos 2\theta = 1- 2\sin^2 \theta.$$ with that you equation becomes $$12(1-2\sin^2 \theta)-6 =\sin \theta \to 24\sin^2 \theta+\sin \theta-6=0, \\ \sin\theta=\frac{-1\pm\sqrt{1+24^2}}{48}= 0.4796,-0.521\\ \theta=28.66^\circ, 151.34^\circ, 211.41^\circ, 328.582^\circ$$ you can add $-360^\circ$ to get the remaining four.