Đáp án:a)\(\left[ \begin{array}{l}x=\frac{-\pi }{24}+k2π\\x=\frac{19\pi }{24}+k2π\end{array} \right.\)
b)\(\left[ \begin{array}{l}x=±arccos(\frac{-1+\sqrt{385}}{24})+k2π\\x=±arccos(\frac{-1-\sqrt{385}}{24})+k2π\end{array} \right.\)
c)\(\left[ \begin{array}{l}x=\frac{π}{2}+k2π\\x=\frac{5π}{6}+k2π\end{array} \right.\)
Giải thích các bước giải:
a)\(sin(2x-\frac{\pi }{3})=\frac{-\sqrt{2}}{2}=sin\frac{-\pi }{4}\)
⇔\(\left[ \begin{array}{l}2x-\frac{\pi }{3}=\frac{-\pi }{4}+k2π\\2x-\frac{\pi }{3}=π+\frac{\pi }{4}+k2π\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{-\pi }{24}+k2π\\x=\frac{19\pi }{24}+k2π\end{array} \right.\)
b) \(6cos^{2}x+cosx-2=0⇔ 6(2cos^{2}x-1)+cosx-2=0⇔12cos^{2}x+cosx-8=0\)
⇔\(\left[ \begin{array}{l}cosx=\frac{-1+\sqrt{385}}{24}\\cosx=\frac{-1-\sqrt{385}}{24}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=±arccos(\frac{-1+\sqrt{385}}{24})+k2π\\x=±arccos(\frac{-1-\sqrt{385}}{24})+k2π\end{array} \right.\)
c)\(\sqrt{3}sinx-cosx=\sqrt{3}\)
⇔\(\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{\sqrt{3}}{2}\)
⇔\(sinxcos\frac{π}{6}-cosxsin\frac{π}{6}=\frac{\sqrt{3}}{2}\)
⇔\(sin(x-\frac{π}{6})=sin\frac{π}{3}\)
⇔\(\left[ \begin{array}{l}x-\frac{π}{6}=\frac{π}{3}+k2π\\x-\frac{π}{6}=π-\frac{π}{3}+k2π\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{π}{2}+k2π\\x=\frac{5π}{6}+k2π\end{array} \right.\)
