Đáp án`+`Giải thích các bước giải:

VD `37:`

`sin2x - 2sinx - 2cosx + 2 =0 `

`<=> 2sinxcosx - 2sinx - 2cosx + 2= 0`

`<=> 2sinx(cosx - 1) - 2(cosx - 1) = 0`

`<=> (cosx - 1)(2sinx - 2) = 0`

`<=>` \(\left[ \begin{array}{l}cosx = 1\\sinx = 1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x= k2\pi\\x= \dfrac{\pi}{2} + k2\pi\end{array} \right.\)  `(k \in ZZ)`

Vậy `S= {k2\pi; (\pi)/2 + k2\pi (k \in ZZ)}`

VD `38:`

`sin7x + sin4x + sinx = 0`

`<=> (sin7x + sinx) + sin4x =0`

`<=> 2sin4xcos3x +sin4x= 0`

`<=> sin4x(2cos3x + 1) = 0`

`<=>` \(\left[ \begin{array}{l}sin4x = 0\\cos3x = \dfrac{-1}{2}\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}4x = k\pi\\3x = \pm \dfrac{2\pi}{3} + k2\pi\end{array} \right.\)  `(k \in ZZ)`

`<=>` \(\left[ \begin{array}{l}x= \dfrac{k\pi}{4}\\x=\pm \dfrac{2\pi}{9} + \dfrac{k2\pi}{3}\end{array} \right.\) `(k \in ZZ)`

Vậy `S= {(k\pi)/4; +- (2\pi)/9 + (k2\pi)/3  (k \in ZZ)}`

`***`sharksosad