cos ² (π /3 +x) +4.cos .(π/6 – x ) = 4 . help meee

Đáp án:

\[x = \dfrac{\pi }{6} + k2\pi \,\,\,\,\,\left( {k \in Z} \right)\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\cos x = \sin \left( {\dfrac{\pi }{2} – x} \right)\\
{\sin ^2}x + {\cos ^2}x = 1\\
{\cos ^2}\left( {\dfrac{\pi }{3} + x} \right) + 4\cos \left( {\dfrac{\pi }{6} – x} \right) = 4\\
 \Leftrightarrow {\sin ^2}\left[ {\dfrac{\pi }{2} – \left( {\dfrac{\pi }{3} + x} \right)} \right] + 4\cos \left( {\dfrac{\pi }{6} – x} \right) = 4\\
 \Leftrightarrow {\sin ^2}\left( {\dfrac{\pi }{6} – x} \right) + 4\cos \left( {\dfrac{\pi }{6} – x} \right) = 4\\
 \Leftrightarrow \left[ {1 – {{\cos }^2}\left( {\dfrac{\pi }{6} – x} \right)} \right] + 4\cos \left( {\dfrac{\pi }{6} – x} \right) = 4\\
 \Leftrightarrow {\cos ^2}\left( {\dfrac{\pi }{6} – x} \right) – 4\cos \left( {\dfrac{\pi }{6} – x} \right) + 3 = 0\\
 \Leftrightarrow \left( {\cos \left( {\dfrac{\pi }{6} – x} \right) – 1} \right)\left( {\cos \left( {\dfrac{\pi }{6} – x} \right) – 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos \left( {\dfrac{\pi }{6} – x} \right) = 1\\
\cos \left( {\dfrac{\pi }{6} – x} \right) = 3\,\,\,\,\,\,\,\,\,\,\,\left( {L,\,\,\, – 1 \le \cos x \le 1} \right)
\end{array} \right.\\
 \Leftrightarrow \cos \left( {\dfrac{\pi }{6} – x} \right) = 1\\
 \Leftrightarrow \dfrac{\pi }{6} – x = k2\pi \\
 \Leftrightarrow x = \dfrac{\pi }{6} + k2\pi \,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)