2 ; 2pi ]...

July 2022 1 2 Report

а) решите уравнение sin^2 x/2 - cos^2 x/2 = cos2x
б) укажите корни уравнения , принадлежащие отрезку [ pi/2 ; 2pi ]

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xERISx

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$\sin^2\dfrac x2-\cos^2 \dfrac x2=\cos (2x)\\\\-\bigg(\cos^2\dfrac x2-\sin^2 \dfrac x2\bigg)=\cos (2x)\\\\-\cos x=\cos (2x)\\-\cos x=2\cos^2x-1\\2\cos^2x+\cos x-1=0\\D=1+4\cdot 2=9=3^2$

$1)~\cos x=\dfrac{-1+3}4=\dfrac12 ~~~~\boxed{\boldsymbol {x_{1,2}=\pm\dfrac{\pi}3+2\pi n,~n\in \mathbb Z}}\\\\ 2)~\cos x=\dfrac{-1-3}4=-1~~~~\boxed{\boldsymbol{x_3=\pi+2\pi k,~k\in \mathbb Z}}$

б)

$x\in\bigg[\dfrac {\pi}2;2\pi\bigg]\\1)~x=\dfrac{\pi}3+2\pi n};~~n=0;~x=\dfrac {\pi}32\pi\\\\2)~x=-\dfrac{\pi}3+2\pi n};~~n=0;~x=-\dfrac {\pi}32\pi$

$3)~x=\pi+2\pi k};~~k=-1;~x=\pi-2\pi=-\pi2\pi$

Ответ : а) $\pm\dfrac{\pi}3+2\pi n;~~\pi+2\pi k;~~n,k\in \mathbb Z$

б) $\dfrac {5\pi}3\in\bigg[\dfrac {\pi}2;2\pi\bigg];~~\pi\in\bigg[\dfrac {\pi}2;2\pi\bigg]$

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