Question

4cos^4x-3cos2x-1=0 объясните!пожалуйста

Answer 1

$4cos^4x-3cos2x-1=0\\\\4cos^4x-3(2cos^2x-1)-1=0\\\\4cos^4x-6cos^2x+2=0|:2\\\\t=cos^2x \geq 0,\; \; 2t^2-3t+1=0\\\\D=9-8=1,t_1=\frac{1}{2},\; t_2=1\\\\cos^2x=\frac{1+cos2x}{2}=\frac{1}{2},\; \to \; cos2x=0;\; 2x=\frac{\pi}{2}+\pi n,\; x=\frac{\pi}{4}+\frac{\pi n}{2},n\in Z\\\\cos^2x=\frac{1+cos2x}{2}=1,\; cos2x=1,\; 2x=2\pi k,\; x=\pi k,\; k\in Z$