Giải phương trình: cos(3x + pi/3) = -1/2
Giải phương trình:
\(\cos \left( {3x + \frac{\pi }{3}} \right) = - \frac{1}{2}\);
Sử dụng định lí tổng 3 góc trong tam giác.
Do A + C = π – B nên sin(A + C) = sin(π – B) = sin B.
Vậy sin B = sin(A + C
Do \(\cos \frac{{2\pi }}{3} = - \frac{1}{2}\) nên \(\cos \left( {3x + \frac{\pi }{3}} \right) = - \frac{1}{2}\)\( \Leftrightarrow \cos \left( {3x + \frac{\pi }{3}} \right) = \cos \frac{{2\pi }}{3}\)
\( \Leftrightarrow \left[ \begin{array}{l}3x + \frac{\pi }{3} = \frac{{2\pi }}{3} + k2\pi \\3x + \frac{\pi }{3} = - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\)
\( \Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{3} + k2\pi \\3x = - \pi + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\)
\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{9} + k\frac{{2\pi }}{3}\\x = - \frac{\pi }{3} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\).
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