solving for the left side for a double angle formula $\cos t-\sin 2t=0$
Solve the left hand side so that it equals zero.
Do I use $(2\sin t \cos {t})$ for $\sin 2t$?
 A: Sure, then $$\cos t - \sin(2t) = \cos t - 2\sin t \cos t = \cos t(1 - 2\sin t) = 0$$
Now, $\cos t(1-2\sin t) = 0$ if and only if either 


*

*$\cos t = 0 \implies t = ?$, or else 

*$1 - 2\sin t = 0 \iff 2\sin t = 1 \iff \sin t = 1/2$.
For what values of $t$ does $\cos t = 0\;\;?$ For example, we know that $\cos (\pi/2) = 0$, and $\cos (3\pi/2) = 0$.
For what values of $t$ does $\sin t = 1/2\;\;?$ Recall that $\sin(\pi/6) = 1/2$. Consider what other angles, $t$ make $\sin t = 1/2$.
A: Yes, you can use the double angle formula.
$$\cos{t} - \sin{2t} = 0\\
\cos{t} - 2\sin{t}\cos{t} = 0\\
\cos{t}(1 - 2\sin{t}) = 0\\$$
From this, $\cos{t} = 0$ or $\sin{t} = \frac{1}{2}$. Now, just solve for $t$ for both cases.
A: Not sure why you need "Double Angle" Formula
As $\displaystyle\sin2t=\cos t=\sin\left(\frac\pi2-t\right)$
$\displaystyle\implies2t=n\pi+(-1)^n t$ where $n$ is any integer 
Now check for the odd & the even cases

Alternatively, $\displaystyle\cos t=\sin2t=\cos\left(\frac\pi2-2t\right)$
$\displaystyle\implies t=2m\pi\pm\left(\frac\pi2-2t\right) $ where $m$ is any integer 
Now check for either signs separately.
