Giải pt. 1) tan(x - bi/3)=cot2x 2) cos^2x= 1 + sin7x

Đáp án:

\(1)\,\,x = \frac{{5\pi }}{{18}} + \frac{{l\pi }}{3}\,\,\,\left( {l \in Z} \right)\)

Giải thích các bước giải:

\(\begin{array}{l}
1)\,\,\tan \left( {x - \frac{\pi }{3}} \right) = \cot \left( {2x} \right)\,\,\,\,\left( * \right)\\
DK:\,\,\,\left\{ \begin{array}{l}
\cos \left( {x - \frac{\pi }{3}} \right) \ne 0\\
\sin \left( {2x} \right) \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x - \frac{\pi }{3} \ne \frac{\pi }{2} + k\pi \\
2x \ne m\pi 
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \frac{{5\pi }}{6} + k\pi \\
x \ne \frac{{m\pi }}{2}
\end{array} \right.\,\,\,\left( {k,\,\,m \in Z} \right).\\
\left( * \right) \Leftrightarrow \tan \left( {x - \frac{\pi }{3}} \right) = \tan \left( {\frac{\pi }{2} - 2x} \right)\\
 \Leftrightarrow x - \frac{\pi }{3} = \frac{\pi }{2} - 2x + l\pi \\
 \Leftrightarrow x = \frac{{5\pi }}{{18}} + \frac{{l\pi }}{3}\,\,\,\left( {l \in Z} \right)\\
 \Rightarrow pt\,\,co\,\,nghiem \Leftrightarrow \left\{ \begin{array}{l}
\frac{{5\pi }}{{18}} + \frac{{l\pi }}{3} \ne \frac{{5\pi }}{6} + k\pi \\
\frac{{5\pi }}{{18}} + \frac{{l\pi }}{3} \ne \frac{{m\pi }}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
l \ne \frac{5}{3} + 3k\\
l \ne  - \frac{5}{6} + \frac{{3m}}{2}
\end{array} \right.\,\,\,tm\,\,\forall m,\,\,l,\,\,k \in Z.
\end{array}\)

Đề bài câu 2 của bạn là: \({\cos ^2}x = 1 + \sin 7x\) à bạn?