Question

2sinxcos2x-1+2cos2x-sinx=0

Answer 1

$-\dfrac{\pi }{2}+2\pi k,~k\in\mathbb {Z};\pm\dfrac{\pi }{6}+\pi n,~n\in\mathbb {Z}$

Объяснение:

$2sinxcos2x -1+2cos2x-sinx=0;\\(2sinxcos2x +2cos2x)-(1+sinx)=0;\\2cos2x(sinx+1)-(sinx+1)=0;\\(sinx+1)(2cos2x-1)=0;$

$\left [\begin{array}{l} sinx +1= 0,\\ 2cos2x-1 = 0; \end{array} \right.\Leftrightarrow \left [\begin{array}{l} sinx = -1,\\ cos2x= \dfrac{1}{2} ; \end{array} \right.\Leftrightarrow \left [\begin{array}{l} x= -\dfrac{\pi }{2}+2\pi k,~k\in\mathbb {Z},\\\\2x= \pm\dfrac{\pi }{3}+2\pi n,~n\in\mathbb {Z} ; \end{array} \right.\Leftrightarrow$

$\Leftrightarrow \left [\begin{array}{l} x= -\dfrac{\pi }{2}+2\pi k,~k\in\mathbb {Z},\\\\x= \pm\dfrac{\pi }{6}+\pi n,~n\in\mathbb {Z} . \end{array} \right.$