Question

80 балов
Решите уравнение
4 cos²⁡x − sin²⁡x = sin⁡ x

Answer 1

$4\cos ^2x-\sin^2x=\sin x$

$4(1-\sin^2x)-\sin^2x=\sin x$

$4-4\sin^2x-\sin^2x-\sin x=0$

$5\sin^2x+\sin x-4=0$

$5\sin^2x+5\sin x-4\sin x-4=0$

$5\sin x(\sin x+1)-4(\sin x+1)=0$

$(\sin x+1)(5\sin x-4)=0$

$\sin x+1=0$

$x=-\dfrac{\pi}{2}+2\pi k,k \in Z$

$5\sin x-4=0$

$x=(-1)^k\cdot \arcsin\frac{4}{5}+\pi m, m \in Z$

Answer 2

$\displaystyle\bf\\4Cos^{2} x-Sin^{2} x=Sinx\\\\4\cdot(1-Sin^{2} x)-Sin^{2} x-Sinx=0\\\\4-4Sin^{2} x-Sin^{2}x-Sinx=0\\\\5Sin^{2}x+Sinx-4=0\\\\Sinx=m \ , \ -1\leq m\leq 1\\\\5m^{2} +m-4=0\\\\D=1^{2} -4\cdot 5\cdot(-4)=1+80=81=9^{2} \\\\m_{1} =\frac{-1-9}{10} =-1\\\\\\m_{2} =\frac{-1+9}{10} =0,8\\\\1)\\\\Sinx=-1\\\\x=-\frac{\pi }{2} +2\pi n,n\in Z\\\\2)\\\\Sinx=0,8\\\\x=(-1)^{n} arcSin0,8+\pi n,n\in Z\\\\Otvet:-\frac{\pi }{2} +2\pi n \ ; \ (-1)^{n }arcSin0,8+\pi n,n\in Z$