1.cosx + sin(2x + 3 π/4)=0 2.cos(2x + π/3)+ cos (x – π/6)=0 3. tan3x + tan (2x – π/4) = 0

0 bình luận về “1.cosx + sin(2x + 3 π/4)=0 2.cos(2x + π/3)+ cos (x – π/6)=0 3. tan3x + tan (2x – π/4) = 0”

1. 1) $\cos x + \sin\left(2x +\dfrac{3\pi}{4}\right) = 0$

   $\Leftrightarrow \cos x = – \sin\left(2x +\dfrac{3\pi}{4}\right)$

   $\Leftrightarrow \cos x = \cos\left(\dfrac{\pi}{2} + x +\dfrac{3\pi}{4}\right)$

   $\Leftrightarrow \left[\begin{array}{l}x = x + \dfrac{5\pi}{4} + k2\pi\\x = – x – \dfrac{5\pi}{4} + k2\pi\end{array}\right.$

   $\Leftrightarrow x = -\dfrac{5\pi}{8} + k\pi \quad (k \in \Bbb Z)$

   2) $\cos\left(2x + \dfrac{\pi}{3}\right) + \cos\left(x – \dfrac{\pi}{6}\right) = 0$

   $\Leftrightarrow \cos\left(2x + \dfrac{\pi}{3}\right) = -\cos\left(x – \dfrac{\pi}{6}\right)$

   $\Leftrightarrow \cos\left(2x + \dfrac{\pi}{3}\right) = \cos\left(\pi – x + \dfrac{\pi}{6}\right)$

   $\Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{3} = -x + \dfrac{7\pi}{6} + k2\pi\\2x + \dfrac{\pi}{3}= x – \dfrac{7\pi}{6} + k2\pi\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}x + \dfrac{5\pi}{18} + k\dfrac{2\pi}{3}\\x =\dfrac{3\pi}{2} + k2\pi\end{array}\right.$

   3) $\tan3x + \tan\left(2x -\dfrac{\pi}{4}\right) = 0\quad (*)$

   $ĐK: \, \begin{cases}\cos3x \ne 0\\\cos\left(2x -\dfrac{\pi}{4}\right) \ne 0\end{cases}\Leftrightarrow \begin{cases}x\ne \dfrac{\pi}{6} + k\dfrac{\pi}{3}\\x \ne \dfrac{\pi}{8} + k\dfrac{\pi}{2}\end{cases}\quad (k\in \Bbb Z)$

   $\Leftrightarrow \tan3x = \tan\left(\dfrac{\pi}{4}-2x\right) = 0$

   $\Leftrightarrow 3x = \dfrac{\pi}{4} – 2x + k\pi$

   $\Leftrightarrow x = \dfrac{\pi}{20} + k\dfrac{\pi}{5}\quad (k\in \Bbb Z)$

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