cos2x-sinx=0решите пожалуйста...

mamapopova199 @mamapopova199

August 2022 1 9 Report

cos2x-sinx=0решите пожалуйста

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Ğåłåxý Cos2x= 1-2sin^2x
$1 - 2\sin ^{2} (x) - \sin(x) = 0 \\ \sin(x) = y \\ 1 - 2y {}^{2} - y = 0 \\ 2y {}^{2} + y - 1 = 0 \\ d = 1 + 8 = 3 { }^{2} \\ x = \frac{ - 1 + 3}{4} = \frac{1}{2} \\ x = \frac{ - 1 - 3}{4} = - 1 \\ \sin(x) = \frac{1}{2} \\ x = \frac{\pi}{6} + 2\pi \: n \: \\ x = \frac{5 \pi}{6} + 2\pi \: n \\ \\ sin(x) = - 1 \\ x = - \frac{\pi}{2} + 2\pi \: k$

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