Question

ПОМОГИТЕ РЕШИТЬ УРАВНЕНИЕ УМОЛЯЮ 4sin^2x-3cosx-3=0

Answer 1

$\displaystyle\bf\\4Sin^{2} x-3Cosx-3=0\\\\4\cdot(1-Cos^{2} x)-3Cosx-3=0\\\\4-4Cos^{2}x-3Cosx-3=0\\\\4Cos^{2} x+3Cosx-1=0\\\\Cosx=m \ , \ \ -1\leq m\leq 1\\\\4m^{2} +3m-1=0\\\\D=3^{2} -4\cdot 4 \cdot (-1)=9+16=25=5^{2} \\\\m_{1} =\frac{-3-5}{8} =-1\\\\m_{2} =\frac{-3+5}{8} =\frac{1}{4} \\\\1)\\Cosx=-1\\\\x=\pi +2\pi n,n\in Z\\\\2)\\Cosx=\frac{1}{4} \\\\x=\pm \ arcCos\frac{1}{4} +2\pi n,n\in Z$