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: $(a)\sin2x=2\sin x\cos x$
$\implies\sin2x=-1=-\sin\dfrac\pi2=\sin\left(-\dfrac\pi2\right)$
$2x=m\pi+(-1)^m\left(-\dfrac\pi2\right)$ 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 $\cos x<0$
A: The double angle formulas for sine, cosine, and tangent are 
\begin{align*}
\sin(2x) & = 2\sin x\cos x\\
\cos(2x) & = \cos^2x - \sin^2x\\
         & = 2\cos^2x - 1\\
         & = 1 - 2\cos^2x\\
\tan(2x) & = \frac{2\tan x}{1 - \tan^2x}
\end{align*}
Of course, if you know $\sin(2x)$ and $\cos(2x)$, you can compute $\tan(2x)$ by dividing $\sin(2x)$ by $\cos(2x)$.  
For the first problem:
\begin{align*}
8\sin x\cos x + 4 & = 0\\
2\sin x\cos x + 1 & = 0\\
\sin(2x) + 1 & = 0 && \text{using the identity $\sin(2x) = 2\sin x\cos x$}\\
\sin(2x) & = -1\\
2x & = -\frac{\pi}{2} + 2n\pi, n \in \mathbb{Z}\\
x & = -\frac{\pi}{4} + n\pi, n \in \mathbb{Z}
\end{align*}
Since $x \in (0, 4\pi]$, $n = 1, 2, 3, 4$, which yields the solutions
$x = \dfrac{3\pi}{4}, \dfrac{7\pi}{4}, \dfrac{11\pi}{4}, \dfrac{15\pi}{4}$.  
You can verify that these solutions are correct by direct substitution.  For instance, 
$$8\sin\left(\frac{3\pi}{4}\right)\cos\left(\frac{3\pi}{4}\right) + 4 = 8\left(\frac{\sqrt{2}}{2}\right)\left(-\frac{\sqrt{2}}{2}\right) + 4 = -4 + 4 = 0$$
For the second problem:
Since $\frac{\pi}{2} < x < \pi$, $x$ is a second-quadrant angle, so $\cos x < 0$.  You know the value of $\sin x$.  To use the double angle formulas for sine and cosine, you must determine $\cos x$.  One option would be to draw a right triangle in the second quadrant with opposite side of length $5$ and hypotenuse of length $7$, then use the triangle to determine the cosine.  Otherwise, you can use the Pythagorean Identity $\sin^2x + \cos^2x = 1$ to solve for $\cos x$, remembering to take the negative root.  Once you find $\cos x$, substitute the values of $\sin x$ and $\cos x$ into the double angle formulas for $\sin(2x)$ and $\cos(2x)$, from which you can determine the values of the other trigonometric functions.
A: Use law of sines and cosines to find $\cos 2x$ and $\sin 2x$. Note that $$\tan 2x = \frac{\sin 2x}{\cos 2x}$$

A: Here are the solutions.
A) $4\cdot\sin2x=-4$. This means $\sin2x=-1$. $\sin$ takes value $-1$ at $3\pi/2$ and $7\pi/2$ in the given domain. This means that $2x=3\pi/2$, implying $x=3\pi/4$ and similarly $2x=7\pi/2$ , therefore $x=7\pi/4$.
For b) part, just find the formulates for trigonometric double angle and since $x$ is greater than $\pi/2$ and smaller than $\pi$, $\sin$ will be positive but rest all will be negative.
