Question

Sin5x+sinx+2sin^2x=1    

Answer 1

$></p><p><img src=$ 0 0

Answer 2

Ответ:

$\frac{\pi }{4} +\frac{\pi n}{2} , (-1)^{k} *\frac{\pi }{18} +\frac{\pi k}{3} ,~n,k\in\mathbb {Z}.$

Объяснение:

$sin5x+sinx+2sin^{2}x =1;\\2sin\frac{5x+x}{2} *cos \frac{5x-x}{2} =1- 2sin^{2} x;\\2sin3x*cos2x=cos2x;\\2sin3x*cos2x-cos2x=0;\\cos2x(2sin3x-1)=0;\\\left [ \begin{array}{lcl} {{cos2x=0,} \\ {sin3x=\frac{1}{2} ;}} \end{array} \right.\Leftrightarrow\left [ \begin{array}{lcl} {{2x=\frac{\pi }{2}+\pi n,~n\in\mathbb {Z} } \\ {3x=(-1)^{k}*\frac{\pi }{6} +\pi k , ~k\in\mathbb {Z}}} \end{array} \right.\Leftrightarrow$

$\left [ \begin{array}{lcl} {{x=\frac{\pi }{4} +\frac{\pi n}{2} ~n\in\mathbb {Z},} \\ {x=(-1)^{k} *\frac{\pi }{18}+\frac{\pi k}{3} ,~k\in\mathbb {Z} .}} \end{array} \right.$