Double-Angle or Half-Angle to solve equation from [0,2pi) I am currently taking a Pre Calc II (Trig) class in college. I have a problem that I do not know how to solve and was wondering if I could get some help.  I have done some work but have gotten stuck. Here is the problem:
Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2π). 
$-\sin(2\theta)-\cos(4\theta)=0$
Here is what I have done: 
$-\sin(2\theta)-2\cos^2(2\theta)-1$
$-2\sin(\theta)\cos(\theta)-2\cos^2(2\theta)-1=0$
After this point, I do not know how to continue so maybe I made a mistake with expanding one of the identities? 
Thanks in advance!
 A: using the identity that @rogerl provided you have: $$ sin(2\theta) +1-2sin^2(2\theta)=0$$
which when you substitute $sin(2\theta) = \alpha$ becomes a quadratic equation: $$2\alpha^2-\alpha-1=0$$
you can solve the following and equate the values for $\alpha$ to $sin(2\theta)$ would yield the desired results i would persume.
A: HINT:
Can you figure out from the short hand?
$$ S_{2t}+ C_{4t} =0 $$
$$ S+1-2S^2=0 \rightarrow \, S= (1,-\frac12)$$
$$S_{2t}=1\rightarrow 2t= \pi/2, 5 \pi/2,... $$
$$ S_{2t}=-\frac12 \rightarrow 2t= 7 \pi/6, 11\pi/6,..$$
shall explain if required.
A: I'll set $\phi=\theta$ for convenience. So your equation is equivalent to $$\sin2\phi+\sin(π/2-4\phi)=0.$$ This becomes $$2\sin\frac12\left(2\phi+\fracπ2-4\phi\right)\cos\frac12\left(2\phi-\fracπ2+4\phi\right)=0,$$ or more simply $$\sin\left(\fracπ4-\phi\right)\cos\left(3\phi-\fracπ4\right)=0.$$ Thus we have that $$\sin\left(\fracπ4-\phi\right)=0,$$ or that $$\cos\left(3\phi-\fracπ4\right)=0.$$ It follows that  $$\fracπ4-\phi=πj$$ or $$3\phi-\fracπ4=(1+2k)\fracπ2,$$ where $j,k$ are arbitrary integers. Now solve for $\phi$ in each of these equations and impose the conditions $$0\le\phi\lt 2π$$ to find your solutions.
