Finding $\sin 2x, \cos 2x, \tan 2x$, & finding all solutions of $x$? I am a senior in high school at the moment taking a pre-calculus course. I was out a few days this week and last due to illness and I have a big test tomorrow. Please help me figure out how to do these two problems on my review sheet, and please show me the steps!
a) $8\sin x\cos x + 4 = 0$ , find all solutions for $x$ in the interval $(0, 4\pi]$ 
b) $\sin x = \frac{5}{7}$ , find $\sin 2x, \cos 2x$, and $\tan 2x$
cosine is negative since it's between $\pi$ and $\frac{\pi}{2}$.

$(a)\sin2x=2\sin x\cos x$
$\implies\sin2x=-1=-\sin\dfrac\pi2=\sin\left(-\dfrac\pi2\right)$
$2x=m\pi+(-1)^m\left(-\dfrac\pi2\right)$ where $m$ is any integer
For even $m,m=2n\implies2x=2n\pi-\dfrac\pi2=(4n-1)\dfrac\pi2$
For odd $m,m=2n+1\implies2x=(2n+1)\pi+\dfrac\pi2=(4n+3)\dfrac\pi2$
Observe that both cases represent actually the same set
We need $0\le(4n-1)\dfrac\pi2<4\pi\iff0\le4n-1<8\iff.25\le n<2.25\implies n=1,2$
$(b)$  See Double-Angle Formulas  and use $\cos^2x=1-\sin^2x$ and $\cos x<0$
