Câu hỏi:
Giải phương trình \(\cos 2x + \sin 2x = \sqrt 2 \cos x\) .
-
A.
\(\left[ \begin{array}{l}x = \dfrac{\pi }{4} + k\dfrac{{2\pi }}{3}\\x = \dfrac{{3\pi }}{4} + k2\pi \end{array} \right.\) -
B.
\(\left[ \begin{array}{l}x = – \dfrac{\pi }{4} + k2\pi \\x = – \dfrac{\pi }{{12}} + k\dfrac{{2\pi }}{3}\end{array} \right.\) -
C.
\(\left[ \begin{array}{l}x = \dfrac{\pi }{4} + k2\pi \\x = \dfrac{{4\pi }}{9} + k\dfrac{{2\pi }}{3}\end{array} \right.\) -
D.
\(\left[ \begin{array}{l}x = \dfrac{\pi }{4} + k2\pi \\x = \dfrac{\pi }{{12}} + k\dfrac{{2\pi }}{3}\end{array} \right.\)
Lời giải tham khảo:
Đáp án đúng: D
Ta có: \(\cos 2x + \sin 2x = \sqrt 2 \cos x\)
\(\begin{array}{l}
\cos 2x + \sin 2x = \sqrt 2 \cos x\\
\Leftrightarrow \sqrt 2 \cos \left( {2x – \frac{\pi }{4}} \right) = \sqrt 2 \cos x\\
\Leftrightarrow \cos \left( {2x – \frac{\pi }{4}} \right) = \cos x\\
\Leftrightarrow \left[ \begin{array}{l}
2x – \frac{\pi }{4} = x + k2\pi \\
2x – \frac{\pi }{4} = – x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{4} + k2\pi \\
3x = \frac{\pi }{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{4} + k2\pi \\
x = \frac{\pi }{{12}} + \frac{{k2\pi }}{3}
\end{array} \right.
\end{array}\)
Chọn đáp án D.
Trả lời