Question

Розв'язати нерівність:
$sin^{4}x-cos^{4} x\geq \sqrt{3} /2$

Answer 1

$\displaystyle\bf\\Sin^{4} x-Cos^{4} x\geq \frac{\sqrt{3} }{2} \\\\\\\underbrace{(Sin^{2} x+Cos^{2} x)}_{1}\cdot\underbrace{(Sin^{2} x-Cos^{2} x)}_{-Cos2x}\geq \frac{\sqrt{3} }{2} \\\\\\-Cos2x\geq \frac{\sqrt{3} }{2} \\\\\\Cos2x\leq -\frac{\sqrt{3} }{2} \\\\\\\frac{5\pi }{6}+2\pi n\leq 2x\leq \frac{7\pi }{6} +2\pi n,n\in Z\\\\\\\frac{5\pi }{12}+\pi n\leq x\leq \frac{7\pi }{12} +\pi n,n\in Z\\\\\\Otvet \ : \ x\in\Big[\frac{5\pi }{12} +\pi n \ ; \ \frac{7\pi }{12} +\pi n\Big] \ ,n\in Z$