Question

Решите тригонометрическое равнение
$\cos(2x) - 1 = \sqrt{2} \sin( \frac{5\pi}{2} - x )$

Answer 1

$cos(2x)-1= \sqrt{2} sin(\frac{5\pi}{2} -x)\\cos^2(x)-sin^2(x)-1= \sqrt{2} cos(x)\\cos^2(x)-sin^2(x)-(sin^2(x)+cos^2(x))= \sqrt{2} cos(x)\\cos^2(x)-sin^2(x)-sin^2(x)-cos^2(x)= \sqrt{2} cos(x)\\-2sin^2(x)= \sqrt{2} cos(x)\\-2sin^2(x)- \sqrt{2}cos(x)=0\\-2+2cos^2(x)- \sqrt{2} cos(x)=0\\cos(x)=t,-1 \leq t \leq 1\\-2+2t^2- \sqrt{2} t=0\\D=18\\t_1=\frac{ \sqrt{2}+ \sqrt{18} }{4} = \sqrt{2} \\t_2-\frac{ \sqrt{2} }{2} \\cos(x)=-\frac{ \sqrt{2} }{2} \\x=+-arccos(-\frac{ \sqrt{2} }{2} )+2\pi k\\x=+-\frac{3\pi}{4} +2\pi k$