Đáp án:
\(\begin{array}{l}
e)\quad S = \left\{-\dfrac{\pi}{12} + k\pi;\ \dfrac{7\pi}{12} + k\pi;\ \dfrac{\pi}{4} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}\\
f)\quad S = \left\{\dfrac{5\pi}{36} + k\dfrac{\pi}{3}\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
e)\quad \sin\left(4x + \dfrac{\pi}{6}\right) + \cos\left(2x - \dfrac{\pi}{6}\right) = 0\\
\Leftrightarrow \dfrac{\sqrt3}{2}\sin4x + \dfrac12\cos4x + \dfrac{\sqrt3}{2}\cos2x + \dfrac12\sin2x = 0\\
\Leftrightarrow 2\sqrt3\sin2x\cos2x + 1 - 2\sin^22x + \sqrt3\cos2x + \sin2x =0\\
\Leftrightarrow \sqrt3\cos2x(2\sin2x + 1) + (\sin2x- 1)(2\sin2x + 1) = 0\\
\Leftrightarrow (2\sin2x + 1)(\sin2x + \sqrt3\cos2x -1) = 0\\
\Leftrightarrow \left[\begin{array}{l}\sin2x = - \dfrac12\\\sin2x + \sqrt3\cos2x = 1\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}\sin2x = - \dfrac12\\\sin\left(2x + \dfrac{\pi}{3}\right) = \dfrac12\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}2x = -\dfrac{\pi}{6} + k2\pi\\2x = \dfrac{7\pi}{6} + k2\pi\\2x + \dfrac{\pi}{3} = \dfrac{\pi}{6} + k2\pi\\2x + \dfrac{\pi}{3} = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{12} + k\pi\\x = \dfrac{7\pi}{12} + k\pi\\x = \dfrac{\pi}{4} + k\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{-\dfrac{\pi}{12} + k\pi;\ \dfrac{7\pi}{12} + k\pi;\ \dfrac{\pi}{4} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}\\
f)\quad \cot\left(2x - \dfrac{\pi}{4}\right) =\tan\left(x+ \dfrac{\pi}{3}\right)\qquad (*)\\
ĐK:\begin{cases}\sin\left(2x - \dfrac{\pi}{4}\right) \ne 0\\\cos\left(x+ \dfrac{\pi}{3}\right) \ne 0\end{cases}\\
\Leftrightarrow \begin{cases}x \ne \dfrac{\pi}{8} + \dfrac{n\pi}{2}\\x \ne \dfrac{\pi}{6} + n\pi\end{cases}\\
(*)\Leftrightarrow \tan\left(\dfrac{3\pi}{4}-2x\right) =\tan\left(x+ \dfrac{\pi}{3}\right)\\
\Leftrightarrow \dfrac{3\pi}{4} - 2x = x + \dfrac{\pi}{3} + k\pi\\
\Leftrightarrow x = \dfrac{5\pi}{36} + k\dfrac{\pi}{3}\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{\dfrac{5\pi}{36} + k\dfrac{\pi}{3}\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)
