Đáp án:
2. \(\left[ \begin{array}{l}x=-\dfrac{17\pi}{108}+\dfrac{k2\pi}{3}\\x=\dfrac{49\pi}{108}+\dfrac{k2\pi}{3}\end{array} \right.\)
Giải thích các bước giải:
1. $sin x= 1-2sinx^2$
⇔ $2sinx^2+ sin x-1=0⇔$ \(\left[ \begin{array}{l}sin x=\dfrac{1}{2}\\sin x=-1\end{array} \right.\)
Với $sin x= \dfrac{1}{2}⇒$ \(\left[ \begin{array}{l}x=\dfrac{\pi}{6}+ k2\pi\\x=\dfrac{5\pi}{6}+ k2\pi\end{array} \right.\)
Với $sin x= -1$ ⇒\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+ k2\pi\\x=\dfrac{3\pi}{2}+ k2\pi\end{array} \right.\)
2. $sin ( 3x+ \dfrac{5\pi}{36})=sin (-\dfrac{\pi}{3})$
⇔\(\left[ \begin{array}{l}3x+\dfrac{5\pi}{36}=-\dfrac{\pi}{3}+ k2\pi\\3x+\dfrac{5\pi}{36}=\dfrac{4\pi}{3}+k2\pi\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-\dfrac{17\pi}{108}+\dfrac{k2\pi}{3}\\x=\dfrac{49\pi}{108}+\dfrac{k2\pi}{3}\end{array} \right.\)
3. $sin x= sin (\dfrac{\pi}{2}-\dfrac{1}{x})$
\(\left[ \begin{array}{l}x= \dfrac{\pi}{2}-\dfrac{1}{x}+k2\pi\\x=\dfrac{\pi}{2}+\dfrac{1}{x}+ k2\pi\end{array} \right.\)
