<s>[INST] 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 $\frac{\pi}{2}$.
 [/INST] 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.
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