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Question Number 186685 by norboyev last updated on 08/Feb/23

$${\left(\sin x\right)}^2-\frac{1}{2}\sin 2x-2{\left(\cos x\right)}^2\ge 0$$$$x\in [0;2\pi ]$$

Answered by Ar Brandon last updated on 08/Feb/23

$${\sin }^{2}x-\frac{1}{2}\sin 2x-{2\cos }^{2}x⩾0$$$$\Rightarrow {\sin }^{2}x-\sin x\cos x-{2\cos }^{2}x⩾0$$$$\Rightarrow s^2-s\sqrt{1-s^2}-2(1-s^2)⩾0(\mid s\mid ⩽1)$$$$\Rightarrow 3s^2-2⩾s\sqrt{1-s^2}\Rightarrow 9s^4-12s^2+4⩾s^2(1-s^2)$$$$\Rightarrow 10s^4-13s^2+4⩾0\Rightarrow (2s^2-1)(5s^2-4)⩾0$$$$\Rightarrow (\sqrt{2}s-1)(\sqrt{2}s+1)(\sqrt{5}s-2)(\sqrt{5}s+2)⩾0$$$$\Rightarrow s\in (-1;-\frac{2}{\sqrt{5}}]\cup [-\frac{1}{\sqrt{2}};\frac{1}{\sqrt{2}}]\cup [\frac{2}{\sqrt{5}};1)$$$$\Rightarrow x\in [-\frac{\pi }{2};-\arcsin \left(\frac{2}{\sqrt{5}}\right)]\cup [-\arcsin \left(\frac{2}{\sqrt{5}}\right);\arcsin \left(\frac{2}{\sqrt{5}}\right)]$$$$\cup [\arcsin \left(\frac{2}{\sqrt{5}}\right);\frac{\pi }{2}]$$

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