Đáp án:
\(\begin{array}{l}
a)\quad S = \left\{\dfrac{\pi}{8} + k\dfrac{\pi}{4};\ \dfrac{\pi}{2} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}\\
b)\quad S = \left\{-\dfrac{\pi}{12} + k\dfrac{\pi}{2};\ \dfrac{\pi}{12} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}\\
c)\quad S= \left\{- \dfrac{\pi}{6} + k2\pi;\ \dfrac{7\pi}{6} + k2\pi\ \Bigg|\ k\in\Bbb Z\right\}\\
4)\quad S =\left\{\dfrac{\pi}{4} + k\pi;\ \arctan\dfrac14 + k\pi\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\quad \cos5x + \cos3x = 0\\
\Leftrightarrow 2\cos4x.\cos x = 0\\
\Leftrightarrow \left[\begin{array}{l}\cos4x= 0\\\cos x =0\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{8} + k\dfrac{\pi}{4}\\x = \dfrac{\pi}{2} + k\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{\dfrac{\pi}{8} + k\dfrac{\pi}{4};\ \dfrac{\pi}{2} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}\\
b)\quad 4\sin^22x - 3 = 0\\
\Leftrightarrow 4\cdot \dfrac{1 - \cos4x}{2} - 3 = 0\\
\Leftrightarrow 2\cos4x= 1\\
\Leftrightarrow \cos4x = \dfrac12\\
\Leftrightarrow \left[\begin{array}{l}4x = \dfrac{\pi}{3} + k2\pi\\4x = - \dfrac{\pi}{3} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12} + k\dfrac{\pi}{2}\\x = -\dfrac{\pi}{12} + k\dfrac{\pi}{2}\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{-\dfrac{\pi}{12} + k\dfrac{\pi}{2};\ \dfrac{\pi}{12} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}\\
c)\quad \cos2x + 5\sin x + 2 =0\\
\Leftrightarrow (1-2\sin^2x) + 5\sin x +2 = 0\\
\Leftrightarrow 2\sin^2x - 5\sin x -3 =0\\
\Leftrightarrow (2\sin x + 1)(\sin x -3) = 0\\
\Leftrightarrow \left[\begin{array}{l}\sin x = - \dfrac12\\\sin x = 3\quad (vn)\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = - \dfrac{\pi}{6} + k2\pi\\x = \dfrac{7\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S =\left\{- \dfrac{\pi}{6} + k2\pi;\ \dfrac{7\pi}{6} + k2\pi\ \Bigg|\ k\in\Bbb Z\right\}\\
4)\quad 2\sin^2x -5\sin x\cos x - \cos^2x= -2\\
\text{Nhận thấy $\cos x =0$ không là nghiệm của phương trình đã cho}\\
\text{Chia hai vế của phương trình cho $\cos^2x$ ta được:}\\
\quad 2\tan^2x - 5\tan x - 1 = -\dfrac{2}{\cos^2x}\\
\Leftrightarrow 2\tan^2x - 5\tan x - 1 = -2(\tan^2x + 1)\\
\Leftrightarrow 4\tan^2x - 5\tan x + 1 = 0\\
\Leftrightarrow (\tan x -1)(4\tan x -1) = 0\\
\Leftrightarrow \left[\begin{array}{l}\tan x = 1\\\tan x = \dfrac14\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} + k\pi\\x = \arctan\dfrac14 + k\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S =\left\{\dfrac{\pi}{4} + k\pi;\ \arctan\dfrac14 + k\pi\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)
