Giúp mìnhh bài này ạ

Đáp án:

`a)` $\left[\begin{matrix} x= \dfrac{2\pi}{3} + k2\pi \\  x= -\pi + k2\pi\end{matrix}\right.$ `(k \in ZZ)`

`b)`$\left[\begin{matrix} x = -\dfrac{3\pi}{4} + k2\pi\\x = \dfrac{\pi}{12} + \dfrac{k}{3}2\pi\end{matrix}\right.$ `(k \in ZZ)`

`c)` $\left[\begin{matrix} x = -\dfrac{3\pi}{4} + k2\pi\\x = \dfrac{\pi}{12} + \dfrac{k}{3}2\pi\end{matrix}\right.$ `(k \in ZZ)`

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

Giải thích các bước giải:

 `a) 2cos(x + pi/6) = -sqrt{3}`

`<=> cos(x + pi/6) = (-sqrt{3})/2`

`<=>` $\left[\begin{matrix} x+\dfrac{\pi}{6} = \dfrac{5\pi}{6} + k2\pi \\  x+\dfrac{\pi}{6} = -\dfrac{5\pi}{6} + k2\pi\end{matrix}\right.$ `(k \in ZZ)`

`<=>` $\left[\begin{matrix} x= \dfrac{2\pi}{3} + k2\pi \\  x= -\pi + k2\pi\end{matrix}\right.$ `(k \in ZZ)`

`b) sin(2x + pi/4) + cos x = 0`

`<=> sin(2x + pi/4) + sin(pi/2 - x) = 0`

`<=> 2sin\ (2x + pi/4 + pi/2 - x)/2 . sin\(2x + pi/4 - pi/2 + x)/2 = 0`

`<=>` $\left[\begin{matrix} sin(\dfrac{x + \dfrac{3\pi}{4}}{2})= 0 \\ sin(\dfrac{3x -\dfrac{\pi}{4}}{2})= 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} \dfrac{x + \dfrac{3\pi}{4}}{2}= k\pi \\ \dfrac{3x -\dfrac{\pi}{4}}{2}= k\pi\end{matrix}\right.$ `(k \in ZZ)`

`<=>` $\left[\begin{matrix} x = -\dfrac{3\pi}{4} + k2\pi\\x = \dfrac{\pi}{12} + \dfrac{k}{3}2\pi\end{matrix}\right.$ `(k \in ZZ)`

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

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

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

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

`<=>` \(\left[ \begin{array}{l}x = \dfrac{\pi}{2} + k\pi\\sin x = -\dfrac{\sqrt{3}}{2}\end{array} \right.\) 

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

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

`d) sin x + sin 2x + sin 3x = 0`

`<=> (sin x + sin 3x) + sin 2x = 0`

`<=> 2 sin 2x . cos x + sin 2x = 0`

`<=> sin 2x . (2cos x + 1) = 0`

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

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

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