Solve the trigonometric equation/inequality $(1).\quad\cos^2(2x) + \sin^4(x) = 2$ $($solve the equation$)$
$(2).\quad2\cos^2(3x) + 5\cos(3x) - 3 < 0$. 
    For this question, I tried letting $t=\cos(3x)$.
    Thus, $2t^2 + 5t - 3< 0$, but that doesn't factor properly. 
I'm really stuck... It would be nice if someone could show me how to do this step by step... Thanks.
 A: For the equation $$\cos^2(2x) + \sin^4x = 2$$ you can use one of the trigonometric formulas for $\cos(2x)$.  They are
\begin{align*}
\cos^2(2x) & = \cos^2x - \sin^2x\\
          & = 2\cos^2x - 1\\
          & = 1 - 2\sin^2x
\end{align*}
If we use the last one, then we will have a polynomial in $\sin x$.
\begin{align*}
\cos^2(2x) + \sin^4x & = 2\\
[\cos(2x)]^2 + \sin^4x & = 2\\
(1 - 2\sin^2x)^2 + \sin^4x & = 2\\
1 - 4\sin^2x + 4\sin^4x + \sin^4x & = 2\\
5\sin^4x - 4\sin^2x - 1 & = 0\\
5\sin^4x - 5\sin^2x + \sin^2x - 1 & = 0\\
5\sin^2x(\sin^2x - 1) + 1(\sin^2x - 1) & = 0\\
(5\sin^2x + 1)(\sin^2x - 1) & = 0\\
(5\sin^2x + 1)(\sin x + 1)(\sin x - 1) & = 0
\end{align*}
Set the factors equal to zero.  $5\sin^2x + 1 > 0$ for all real numbers $x$.
\begin{align*}
\sin x + 1 & = 0 & \sin x - 1 & = 0\\
\sin x & = -1 & \sin x & = 1\\
x & = -\frac{\pi}{2} + 2n\pi & x & = \frac{\pi}{2} + 2n\pi
\end{align*}
where $n$ is an integer.
Your strategy for the inequality $$2\cos^2(3x) + 5\cos(3x) - 3 < 0$$ is correct.  Factoring yields
\begin{align*}
2\cos^2(3x) + 5\cos(3x) - 3 & < 0\\
2\cos^2(3x) + 6\cos(3x) - \cos(3x) - 3 & < 0\\
2\cos(3x)[\cos(3x) + 3] - 1[\cos(3x) + 3)] & < 0\\
[2\cos(3x) - 1][\cos(3x) + 3] & < 0
\end{align*}
Since $-1 \leq \cos(3x) \leq 1$, $1 \leq \cos(3x) + 3 \leq 4$.  Thus, the term $\cos(3x) + 3$ is always positive.
Therefore, the inequality is satisfied when $2\cos(3x) - 1 < 0$.
\begin{align*}
2\cos(3x) - 1 & < 0\\
2\cos(3x) & < 1\\
\cos(3x) & < \frac{1}{2}
\end{align*}
In the interval $[0, 2\pi)$, the inequality $$\cos u < \frac{1}{2}$$ is satisfied if $$\frac{\pi}{3} < u < \frac{5\pi}{3}$$  In general, the inequality is satisfied if $$\frac{\pi}{3} + 2n\pi < u < \frac{5\pi}{3} + 2n\pi$$  If $u = 3x$, then 
$$\frac{\pi}{9} + \frac{2}{3}n\pi < x < \frac{5\pi}{9} + \frac{2}{3}n\pi$$
A: $$\cos^22x+\sin^2x=2$$
If $\sin^2x=u,\cos2x=1-2u\implies(1-2u)^2+u=2\iff 4u^2-3u-1=0$
$\sin^2x=u=\dfrac{3\pm\sqrt{3^2-4\cdot1(-1)}}8=1,-\dfrac14$
For real $x,\sin^2x\ge0\implies\cos2x=1-2\sin^2x=-1=\cos\pi\implies2x=(2n+1)\pi$ where $n$ is any integer
