Đáp án:
$a$,
\(\left[ \begin{array}{l}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{array} \right.\)
$b$
\(\left[ \begin{array}{l}x=\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{24}+k\pi\end{array} \right.\)
Giải thích các bước giải:
$-2sin^{2}x+3sinx-1=0$
$ đặt $ $sinx=t$ $:$$t∈ [-1;1]$
$pt$ : $-2t^{2}+3t-1=0$
⇔\(\left[ \begin{array}{l}t=1\\t=\frac{1}{2}\end{array} \right.\)
$ với $ $t=1$
⇔ $sin^{}x=1$
⇔$x^{}=\frac{\pi}{2}+k2\pi$
$với $ $ t=$$\frac{1}{2}$
⇔$sin^{}x=$ $\frac{1}{2}$
⇔\(\left[ \begin{array}{l}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{array} \right.\)
$b, \sqrt[]{3}sin2x+cos2x=$ $\sqrt[]{2}$
⇔$\frac{\sqrt3}{2}sin2x+$ $\frac{1}{2}cos2x=$ $\frac{\sqrt2}{2}$
⇔$sin^{}2x.cos$ $\frac{\pi}{6}+cos2x.sin$ $\frac{\pi}{6}=$ $sin\frac{\pi}{4}$
⇔$sin(2x^{}+\frac{\pi}{6})=$ $sin\frac{\pi}{4}$
⇔\(\left[ \begin{array}{l}2x+\frac{\pi}{6}=\frac\pi{4}+k2\pi\\2x+\frac{\pi}{6}=\frac{3\pi}{4}+k2\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{24}+k\pi\end{array} \right.\)
c, $tan(2x-30^{0})=$ $\frac{\sqrt3}{2}$
⇔$tan(2x-\frac{\pi}{6})=^{}$ $\frac{\sqrt3}{2}$
⇔$2x^{}-\frac{\pi}{6}=arctan(\frac{\sqrt3}{2})+k\pi$
⇔$x=\frac{\pi}{12}+arctan(\frac{\sqrt3}{2})+k\pi^{}$
