√3 cos2x + sinx =√3 √3sin2x + cosx =√3

`sqrt{3}cos 2x + sin x = sqrt{3}`

`<=> sqrt{3} - 2sqrt{3}sin^2 x + sin x = sqrt{3}`

`<=> sin x(-2sqrt{3}sin x + 1) = 0`

`<=>` \(\left[ \begin{array}{l}sin x = 0\\sin x = \dfrac{1}{2\sqrt{3}}\end{array} \right.\)

`<=>` \(\left[ \begin{array}{l}x = kπ\\x = arcsin \dfrac{1}{2\sqrt{3}} + k2π\\x = π - arcsin \dfrac{1}{2\sqrt{3}} + k2π\end{array} \right.\) `(k ∈ ZZ)`

`sqrt{3}sin 2x + cos x = sqrt{3}`

`<=> 2sqrt{3}sin x.cos x + cos x = sqrt{3}`

`<=> sqrt{3}cos x(2sin x + 1/(\sqrt{3})cos x) = sqrt{3}`

`<=> cos x(2sin x + 1/(\sqrt{3})cos x) = 0`

`<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{2} + kπ\\2sin x + \dfrac{1}{\sqrt{3}}cos x = 0 (1)\end{array} \right.\) 

Giải `(1)`

`=> 2sqrt{3}sin x + cos x = 0`

`<=> sqrt{13}((2\sqrt{3})/(\sqrt{13})sin x + 1/(\sqrt{13})cos x) = 0`

Đặt

\(\left\{ \begin{array}{l}\dfrac{2\sqrt{3}}{\sqrt{13}} = cos \alpha\\\dfrac{1}{\sqrt{13}} = sin \alpha\end{array} \right.\) 

`=> cos alpha.sin x + sin alpha.cos x = 0`

`<=> sin (x + alpha) = 0`

`<=> x + alpha = kπ`

`<=> x = -alpha + kπ` `(k ∈ ZZ)`