Question Number 186685 by norboyev last updated on 08/Feb/23
![(sinx)^2 −(1/2)sin2x−2(cosx)^2 ≥0 x∈[0;2π]](https://www.tinkutara.com/question/Q186685.png)
$$\left(\mathrm{sin}{x}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}{x}−\mathrm{2}\left(\mathrm{cos}{x}\right)^{\mathrm{2}} \geq\mathrm{0} \\ $$$${x}\in\left[\mathrm{0};\mathrm{2}\pi\right] \\ $$
Answered by Ar Brandon last updated on 08/Feb/23
![sin^2 x−(1/2)sin2x−2cos^2 x≥0 ⇒sin^2 x−sinxcosx−2cos^2 x≥0 ⇒s^2 −s(√(1−s^2 ))−2(1−s^2 )≥0 ( ∣s∣ ≤1 ) ⇒3s^2 −2≥s(√(1−s^2 )) ⇒ 9s^4 −12s^2 +4≥s^2 (1−s^2 ) ⇒10s^4 −13s^2 +4≥0 ⇒ (2s^2 −1)(5s^2 −4)≥0 ⇒((√2)s−1)((√2)s+1)((√5)s−2)((√5)s+2)≥0 ⇒s∈(−1; −(2/( (√5)))]∪[−(1/( (√2))); (1/( (√2)))]∪[(2/( (√5))); 1) ⇒x∈[−(π/2); −arcsin((2/( (√5))))]∪[−arcsin((2/( (√5))));arcsin((2/( (√5))))] ∪[arcsin((2/( (√5)))); (π/2)]](https://www.tinkutara.com/question/Q186687.png)
$$\mathrm{sin}^{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}{x}−\mathrm{2cos}^{\mathrm{2}} {x}\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{2}} {x}−\mathrm{sin}{x}\mathrm{cos}{x}−\mathrm{2cos}^{\mathrm{2}} {x}\geqslant\mathrm{0} \\ $$$$\Rightarrow{s}^{\mathrm{2}} −{s}\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }−\mathrm{2}\left(\mathrm{1}−{s}^{\mathrm{2}} \right)\geqslant\mathrm{0}\:\:\:\left(\:\mid{s}\mid\:\leqslant\mathrm{1}\:\right) \\ $$$$\Rightarrow\mathrm{3}{s}^{\mathrm{2}} −\mathrm{2}\geqslant{s}\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }\:\:\:\:\Rightarrow\:\:\:\mathrm{9}{s}^{\mathrm{4}} −\mathrm{12}{s}^{\mathrm{2}} +\mathrm{4}\geqslant{s}^{\mathrm{2}} \left(\mathrm{1}−{s}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{10}{s}^{\mathrm{4}} −\mathrm{13}{s}^{\mathrm{2}} +\mathrm{4}\geqslant\mathrm{0}\:\:\:\:\Rightarrow\:\:\:\:\left(\mathrm{2}{s}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{5}{s}^{\mathrm{2}} −\mathrm{4}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\left(\sqrt{\mathrm{2}}{s}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}{s}+\mathrm{1}\right)\left(\sqrt{\mathrm{5}}{s}−\mathrm{2}\right)\left(\sqrt{\mathrm{5}}{s}+\mathrm{2}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow{s}\in\left(−\mathrm{1};\:−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right]\cup\left[−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}};\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right]\cup\left[\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}};\:\mathrm{1}\right) \\ $$$$\Rightarrow{x}\in\left[−\frac{\pi}{\mathrm{2}};\:−\mathrm{arcsin}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)\right]\cup\left[−\mathrm{arcsin}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right);\mathrm{arcsin}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)\right] \\ $$$$\:\:\:\:\:\:\cup\left[\mathrm{arcsin}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right);\:\frac{\pi}{\mathrm{2}}\right] \\ $$