Phương trình \({\sin ^2}x + \sqrt 3 {\mathop{\rm s...

D

- Hướng dẫn giải

Ta có: \({\sin ^2}x + \sqrt 3 {\mathop{\rm sinxcosx}\nolimits}  = 1 \Leftrightarrow \frac{{1 - \cos 2x}}{2} + \frac{{\sqrt 3 }}{2}\sin 2x = 1\)

\(\begin{array}{l}
 \Leftrightarrow \frac{{\sqrt 3 }}{2}\sin 2x - \frac{1}{2}\cos 2x = \frac{1}{2} \Leftrightarrow \frac{1}{2}\cos 2x - \frac{{\sqrt 3 }}{2}\sin 2x = \frac{1}{2}\\
 \Leftrightarrow \cos \frac{\pi }{3}.\cos 2x - \sin \frac{\pi }{3}\sin 2x = \frac{1}{2}\\
 \Leftrightarrow \cos \left( {2x + \frac{\pi }{3}} \right) = \cos \frac{\pi }{3} \Leftrightarrow \left[ \begin{array}{l}
2x + \frac{\pi }{3} = \frac{\pi }{3} + k2\pi \\
2x + \frac{\pi }{3} =  - \frac{\pi }{3} + m2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x =  - \frac{\pi }{3} + m\pi 
\end{array} \right.\left( {k,m \in Z} \right)
\end{array}\)

Vì \(x \in \left[ {0;2\pi } \right]\) nên ta có:

\(\begin{array}{l}
0 \le k\pi  \le 2\pi  \Leftrightarrow 0 \le k \le 2 \Leftrightarrow \left[ \begin{array}{l}
k = 0 \Rightarrow x = 0\\
k = 1 \Rightarrow x = \pi \\
k = 2 \Rightarrow x = 2\pi 
\end{array} \right.\\
0 \le  - \frac{\pi }{3} + m2\pi  \le 2\pi  \Leftrightarrow \frac{1}{6} \le m \le \frac{7}{6} \Leftrightarrow m = 1 \Rightarrow x = \frac{{2\pi }}{3}.
\end{array}\)

Vậy có bốn nghiệm thuộc \(\left[ {0;2\pi } \right]\)