Respuesta :

[tex]csc^2x-2cot^2x=0;\ D:x\neq k\pi\ (k\in\mathbb{Z})\\\\\frac{1}{sin^2x}-2\cdot\frac{cos^2x}{sin^2x}=0\\\\\frac{1-2cos^2x}{sin^2x}=0\iff 1-2cos^2x=0\\\\-2cos^2x=-1\ \ \ /:(-2)\\\\cos^2x=\frac{1}{2}\\\\cosx=\sqrt\frac{1}{2}\ \vee\ cosx=-\sqrt\frac{1}{2}\\\\cosx=\frac{\sqrt2}{2}\ \vee\ cosx=-\frac{\sqrt2}{2}[/tex]


[tex]x=\frac{\pi}{4}+2k\pi\ \vee\ x=-\frac{\pi}{4}+2k\pi\ \vee\ x=\frac{3\pi}{4}+2k\pi\ \vee\ x=-\frac{3\pi}{4}+2k\pi\\\\\\Answer:x=\frac{\pi}{4}+\frac{k\pi}{2}\ (k\in\mathbb{Z})[/tex]

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