Giải các phương trình sau:
a) \(sin(x+1)=\frac{2}{3}\)
b) \(sin^22x=\frac{1}{2}\)
c) \(cot^2 \frac{x}{2}=\frac{1}{3}\)
d) \(tan \left ( \frac{x}{12} +12x \right )=-\sqrt{3}\)
Câu a:
\(sin(x+1)=\frac{2}{3}\)
\(\Leftrightarrow \Bigg \lbrack \begin{matrix} x+1 = arcsin \frac{2}{3}+k2 \pi \ \ \ \ \ \\ \\ x+1= \pi -arcsin \frac{2}{3}+k2 \pi \end{matrix}\Leftrightarrow \Bigg \lbrack \begin{matrix} x =-1+ arcsin \frac{2}{3}+k2 \pi \ \ \ \ \ \\ \\ x= -1+\pi -arcsin \frac{2}{3}+k2 \pi \end{matrix}\)
Câu b:
\(sin^22x=\frac{1}{2}\Leftrightarrow sin2x=\pm \frac{1}{\sqrt{2}}\)
* \(sin2x= \frac{1}{\sqrt{2}} \Leftrightarrow sin2x=sin\frac{\pi }{4}\Leftrightarrow \Bigg \lbrack \begin{matrix} 2x=\frac{\pi }{4}+k2\pi \ \ \\ \\ 2x=\frac{3\pi }{4}+k2\pi \ \ \end{matrix}\Leftrightarrow \Bigg \lbrack \begin{matrix} x=\frac{\pi }{8}+k\pi \ \ \\ \\ x=\frac{3\pi }{8}+k\pi \ \ \end{matrix}\)
* \(sin2x=- \frac{1}{\sqrt{2}} \Leftrightarrow sin2x=sin \left ( -\frac{\pi }{4} \right )\Leftrightarrow \Bigg \lbrack \begin{matrix} 2x=-\frac{\pi }{4}+k2\pi \ \ \\ \\ 2x=\frac{5\pi }{4}+k2\pi \ \ \end{matrix}\Leftrightarrow \Bigg \lbrack \begin{matrix} x=-\frac{\pi }{8}+k\pi \ \ \\ \\ x=\frac{5\pi }{8}+k\pi \ \ \end{matrix}\)
Câu c:
\(cot^2\frac{x}{2}=\frac{1}{3}\Leftrightarrow cot \frac{x}{2}=\pm \frac{\sqrt{3}}{3}.\)
* \(cot \frac{x}{2}= \frac{\sqrt{3}}{3}\Leftrightarrow cot\frac{x}{2}=cot\frac{\pi}{3}\Leftrightarrow x=\frac{2\pi }{3}+k2\pi.\)
* \(cot \frac{x}{2}= -\frac{\sqrt{3}}{3}\Leftrightarrow cot\frac{x}{2}=cot\frac{2\pi}{3}\Leftrightarrow x=\frac{4\pi }{3}+k2\pi.\)
Câu d:
\(tan \left ( \frac{\pi }{12} +12x\right )=-\sqrt{3}\)
\(tan \left (12x +\frac{\pi }{12}\right )=tan\frac{2 \pi}{3}\Leftrightarrow 12x +\frac{\pi }{12}= \frac{2 \pi}{3}+k \pi\)
\(\Leftrightarrow x=\frac{7 \pi}{144}+\frac{k \pi}{12}.\)
-- Mod Toán 11
Copyright © 2021 HOCTAP247