<s>[INST] Trigonometric inequality $\sin^{4}x+\cos^{4}x>a$, for $a\in [0,1]$ Solve the inequality: 
$\sin^{4}x+\cos^{4}x>a$, for $a\in [0,1]$
My attempt: 
$(\sin^{2} x+\cos^{2}x)^2-2\sin^{2}x\cos^2{x}>a$
$\Rightarrow 1-2\sin^{2}x\cos^{2}x>a$ 
$\Rightarrow 1-2\sin^{2}x(1-\sin^{2}x)>a$
$\Rightarrow 2\sin^{4}x-2\sin^{2}x+1>a$
$t:=\sin^{2}x$ 
$0\leq 2t^{2}-2t+1\leq 1$ 
$1)$ $2t^{2}-2t\leq 0$ 
$\Rightarrow t\in [0,1] \\$
$2)$ $2t^{2}-2t+1\geq 0$ 
$\Rightarrow t\in \Bbb R$ 
From $1)$ and $2)$ $\Rightarrow t\in [0,1]$ 
$\Rightarrow 0\leq \sin^{2}x \leq 1$ 
$1)$ $\sin^{2}x\leq 1 \Rightarrow -1\leq \sin(x)\leq 1 \Rightarrow x\in \Bbb R$ 
$2)$ $\sin^{2}x\geq 0 \Rightarrow x\in \Bbb R$  
So, $x\in \Bbb R$. 
Is this correct? 
 [/INST] Using $\cos2A=1-2\sin^2A=2\cos^2A-1,$
$$\sin^4x+\cos^4x=\dfrac{(1-\cos2x)^2+(1+\cos2x)^2}4=\dfrac{2+2\cos^22x}4=\dfrac{2+1+\cos4x}4$$
Now $-1\le\cos4x\le1$ for real $x$

Alternatively,
$$\sin^4x+\cos^4x=1-2\sin^2x\cos^2x=1-\dfrac{\sin^22x}2=1-\dfrac{1-\cos4x}4$$
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