Đáp án:

\(\begin{array}{l}
1)\quad S = \left\{x = \dfrac{\pi}{12} + k\pi;\ \dfrac{5\pi}{12} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}\\
2)\quad S = \left\{-\dfrac{\pi}{6} + k2\pi;\ \dfrac{7\pi}{6} + k2\pi\ \Bigg|\ k\in\Bbb Z\right\}\\
3)\quad S = \left\{k\pi;\ \dfrac{\pi}{6} + k2\pi;\ \dfrac{5\pi}{6} + k2\pi\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
1)\quad \tan x + \cot x = 4\qquad (*)\\
ĐK: \begin{cases}\sin x \ne 0\\\cos x \ne 0\end{cases}\Leftrightarrow \sin2x \ne 0 \Leftrightarrow x \ne \dfrac{n\pi}{2}\\
(*)\Leftrightarrow \dfrac{1}{\sin x\cos x} = 4\\
\Leftrightarrow 2\sin x\cos x = \dfrac12\\
\Leftrightarrow \sin2x = \dfrac12\\
\Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{6} + k2\pi\\2x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12} + k\pi\\x = \dfrac{5\pi}{12} + k\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{x = \dfrac{\pi}{12} + k\pi;\ \dfrac{5\pi}{12} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}\\
2)\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\}\\
3)\quad \sin\left(2x + \dfrac{5\pi}{2}\right) -3\cos\left(x - \dfrac{7\pi}{2}\right) = 1 + 2\sin x \\
\Leftrightarrow \sin\left[\left(\dfrac{\pi}{2} + 2x\right) + 2\pi\right] - 3\cos\left[\left(\dfrac{\pi}{2}+x\right)-4\pi\right] = 1 + 2\sin x\\
\Leftrightarrow \sin\left(\dfrac{\pi}{2} + 2x\right) -3\cos\left(\dfrac{\pi}{2} + x\right) = 1 + 2\sin x\\
\Leftrightarrow \cos2x + 3\sin x = 1 +2\sin x\\
\Leftrightarrow (1 - 2\sin^2x) +\sin x- 1 =0\\
\Leftrightarrow 2\sin^2x - \sin x = 0\\
\Leftrightarrow \sin x(2\sin x - 1) = 0\\
\Leftrightarrow \left[\begin{array}{l}\sin x = 0\\\sin x = \dfrac12\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = k\pi\\x = \dfrac{\pi}{6} + k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{k\pi;\ \dfrac{\pi}{6} + k2\pi;\ \dfrac{5\pi}{6} + k2\pi\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)