Đáp án:
Giải thích các bước giải:
$\text{1) $sinx=-\dfrac{\sqrt{2}}{2}$}$ $\\\text{⇔$sinx=sin(-\dfrac{\pi}{4})$}$ $\text{⇔\(\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k2\pi\\x=\pi-(-\dfrac{\pi}{4})+k2\pi\end{array} \right.\)}$ $\\\text{⇔\(\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{5\pi}{4}+k2\pi\end{array} \right.\)$(k∈Z)$}$
$\\\text{2) $sinx=-\dfrac{\sqrt{3}}{2}$}$ $\\\text{⇔$sinx=sin(-\dfrac{\pi}{3})$}$ $\text{⇔\(\left[ \begin{array}{l}x=-\dfrac{\pi}{3}+k2\pi\\x=\pi-(-\dfrac{\pi}{3})+k2\pi\end{array} \right.\)}$ $\\\text{⇔\(\left[ \begin{array}{l}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{4\pi}{3}+k2\pi\end{array}\right.\)$(k∈Z)$}$
$\text{3) }$ $\text{$sinx=sin(\dfrac{\pi}{4})$}$ $\\\text{⇔\(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k2\pi\\x=\pi-\dfrac{\pi}{4}+k2\pi\end{array} \right.\)}$ $\\\text{⇔\(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{3\pi}{4}+k2\pi\end{array} \right.\)$(k∈Z)$}$
