Giải phương trình: a/ √3 sinx-cosx=1 b/ 2sin²x+3sinx–2=0

Đáp án + Giải thích các bước giải:

`a.`

`sqrt3sinx-cosx=1`

`<=>sqrt3/2sinx-1/2cosx=1/2`

`<=>cos\frac{pi}{6}sinx-sin\frac{pi}{6}cosx=sin\frac{pi}{6}`

`<=>sin(x-pi/6)=sin\frac{pi}{6}`

`<=>`\(\left[ \begin{array}{l}x-\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{6}=\pi-\dfrac{\pi}{6}+k2\pi\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}{3}+k2\pi\\x=\pi+k2\pi\end{array} \right.\)`(kinZZ)`

`b.`

`2sin^2x+3sinx–2=0`

`<=>2sin^2x+4sinx-sinx –2=0`

`<=>2sinx(sinx+2)-(sinx+2)=0`

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

`<=>`\(\left[ \begin{array}{l}\sin x=-2 ~~~~\text{(VN)}\\\sin x = \dfrac{1}{2}\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{array} \right.\) `(kinZZ)`