Question

Sin^2x+sin^2п/6=cos^22x+cos^2п/3 решите уравнение

Answer 1

$cos^2x+sin^2\frac{\pi }{6}=cos^22x+cos^2\frac{\pi}{3}\\\\Formyla:\; \; \; \; cos^2x=\frac{1+cos2x}{2}\\\\ \frac{1+cos2x}{2} +(\frac{1}{2})^2=cos^22x+(\frac{1}{2})^2\\\\cos^22x- \frac{1}{2} cos2x-\frac{1}{2} =0\; |\cdot 2\\\\2cos^22x-cos2x-1=0\\\\t=cos2x\; ,\; \; \; -1 \leq t \leq 1\\\\2t^2-t-1=0\\\\D=9\; ,\; \; t_1=-\frac{1}{2}\; ,\; \; t_2=1\\\\a)\; \; cos2x=-\frac{1}{2}\; ,\; \; \\\\2x=\pm (\pi -\frac{\pi}{3})+2\pi n=\pm \frac{2\pi}{3}+2\pi n,\; n\in Z\\\\x=\pm \frac{\pi}{3}+\pi n,\; n\in Z$

$b)\; \; cos2x=1\; ,\; \; 2x=2\pi m\; ,\; m\in Z\\\\x=\pi m\; ,\; m\in Z$