Solving $\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+2}{x^2+2x-2}=\frac52 $ Got these two problems on an exam recently and was unsure if I managed to answer them correctly:
$$\sin^4x + \cos^4x \ge \frac58, \ x \in \left\{-\frac\pi4,\frac\pi4\right\}$$
Second:
$$\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+2}{x^2+2x-2}=\frac52 $$
Anyone able to provide the correct answer or method of solving these?
 A: Hint...for the second one, set $u$ as the first square root expression and solve the quadratic $$u+\frac 1u=\frac 52$$
A: For the first, start from $\sin^2x+\cos^2x=1$ and square both sides.
For the second, note that the second part is the reciprocal of the first. So you need to solve $p+\frac 1p=\frac 52$ first.
But what have you tried?
A: I'll solve both in one shot here using the simplest methods I can think of:  
For the first, we use the proof here
$$\sin^4x + \cos^4x \ge \frac58, \ x \in \left\{-\frac\pi4,\frac\pi4\right\}$$
$$\Rightarrow \displaystyle \frac {\cos (4 x)- 4 \cos (2 x)+3 } 8 + \frac {\cos(4x)+4\cos(2x) + 3} 8 \geq \frac{5}{8}$$
$$\Rightarrow \displaystyle \frac {2\cos (4 x)+6 } 8 \geq \frac{5}{8}$$
$$\Rightarrow \displaystyle 2\cos (4 x)+6 \geq 5$$
$$\Rightarrow \cos(4 x) \geq -\frac{1}{2} $$
$$\Rightarrow 4x \leq \arccos(-1/2) $$
$$\Rightarrow x \leq \frac{1}{4}\bigg(\frac{2\pi}{3}\bigg) $$
$$\Rightarrow x \leq \frac{\pi}{6} $$
Because we used the inverse cosine the domain is restricted to non-negative numbers here, actually making this $0 \leq x \leq \frac{\pi}{6}$. However, since $\cos(x)$ is an even function, we also get $-\frac{\pi}{6} \leq x \leq 0$. Combining these, we get $\color{red}{-\frac{\pi}{6} \leq x \leq \frac{\pi}{6}}$  
For the second problem, $\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+2}{x^2+2x-2}=\frac52$,  we set $u=\sqrt\frac{x^2+2x-2}{x^2+3x+2}$ and, as a result, $\frac{1}{u} = \sqrt\frac{x^2+3x+2}{x^2+2x-2}$. Plugging these into the original equation, we get
$$u + \frac{1}{u} = \frac{5}{2}$$
$$u^2 + 1 = \frac{5}{2}n$$
$$u^2 -\frac{5}{2}u + 1 = 0$$
Solving this, we get $u = 2, \frac{1}{2}$, both of which satisfy our equation in terms of $n$. Substituting, back,
$$\sqrt{\frac{x^2+2x-2}{x^2+3x+2}} = 2 \qquad \text{or} \qquad \sqrt{\frac{x^2+2x-2}{x^2+3x+2}} = \frac{1}{2}$$
$$\frac{x^2+2x-2}{x^2+3x+2} = 4 \qquad \text{or} \qquad \frac{x^2+2x-2}{x^2+3x+2} = \frac{1}{4}$$
The first equation has only complex solutions, but the latter has solutions $\color{red}{x = -\frac{5}{6} \pm \frac{\sqrt{145}}{6}}$, both of which satisfy our original equation in terms of $x$
