Đáp án:

\(\left[ \begin{array}{l}
x =  - \dfrac{\pi }{4} + \arcsin \left( { - \dfrac{{\sqrt 2 }}{4}} \right) + k2\pi \\
x = \dfrac{{3\pi }}{4} - \arcsin \left( { - \dfrac{{\sqrt 2 }}{4}} \right) + k2\pi 
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
3\left( {\sin x + \cos x} \right) = 2.2\sin x.\cos x\left( 1 \right)\\
Đặt:t = \sin x + \cos x\\
 \to {\left( {\sin x + \cos x} \right)^2} = {t^2}\\
 \to {\sin ^2}x + 2\sin x.\cos x + {\cos ^2}x = {t^2}\\
 \to 1 + 2\sin x.\cos x = {t^2}\\
 \to 2\sin x.\cos x = {t^2} - 1\\
\left( 1 \right) \to 3t = 2\left( {{t^2} - 1} \right)\\
 \to 2{t^2} - 3t - 2 = 0\\
 \to \left[ \begin{array}{l}
t = 2\\
t =  - \dfrac{1}{2}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\sin x + \cos x = 2\\
\sin x + \cos x =  - \dfrac{1}{2}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 2\\
\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) =  - \dfrac{1}{2}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\sin \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2 \left( l \right)\\
\sin \left( {x + \dfrac{\pi }{4}} \right) =  - \dfrac{{\sqrt 2 }}{4}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = \arcsin \left( { - \dfrac{{\sqrt 2 }}{4}} \right) + k2\pi \\
x + \dfrac{\pi }{4} = \pi  - \arcsin \left( { - \dfrac{{\sqrt 2 }}{4}} \right) + k2\pi 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - \dfrac{\pi }{4} + \arcsin \left( { - \dfrac{{\sqrt 2 }}{4}} \right) + k2\pi \\
x = \dfrac{{3\pi }}{4} - \arcsin \left( { - \dfrac{{\sqrt 2 }}{4}} \right) + k2\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)

Đáp án: \(\left[ \begin{array}{l}x = -\dfrac{π}{4} + arcsin (-\dfrac{\sqrt{2}}{4}) + k2π\\x = \dfrac{3π}{4} - arcsin (-\dfrac{\sqrt{2}}{4}) + k2π\end{array} \right.\) `(k ∈ ZZ)`

Giải thích các bước giải:

`3(sin x + cos x) = 2sin 2x` 

`<=> 3(sin x + cos x) = 2.2.sin x.cos x`        `(1)`

Đặt `t = sin x + cos x`

`=> 2.sin x.cos x = t² - 1`

`(1)`

`=> 3t = 2t² - 2`

`<=> 2t² - 3t - 2 = 0`

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

`<=> t = -1/2`

`<=> sin x + cos x = -1/2`

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

`<=> sin (x + π/4) = sin (-(\sqrt{2})/4)`

`<=>` \(\left[ \begin{array}{l}x + \dfrac{π}{4} = arcsin (-\dfrac{\sqrt{2}}{4}) + k2π\\x + \dfrac{π}{4} = π - arcsin (-\dfrac{\sqrt{2}}{4}) + k2π\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x = -\dfrac{π}{4} + arcsin (-\dfrac{\sqrt{2}}{4}) + k2π\\x = \dfrac{3π}{4} - arcsin (-\dfrac{\sqrt{2}}{4}) + k2π\end{array} \right.\) `(k ∈ ZZ)`