Giải phương trình tan (pisinx)=0

Đáp án:

Giải thích các bước giải: $\begin{array}{l} \tan \left( {\pi \sin x} \right) = 0 \Leftrightarrow \pi \sin x = k\pi \Leftrightarrow \sin x = k\\ Do\,\, - 1 \le \sin x \le 1 \Rightarrow \left[ \begin{array}{l} k = 0\\ k = 1\\ k = - 1 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} \sin x = 0\\ \sin x = - 1\\ \sin x = 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ x = - \frac{\pi }{2} + 2k\pi \\ x = \frac{\pi }{2} + 2k\pi \end{array} \right. \end{array}$

ĐK: $\cos(\pi.\sin x)\neq 0$

$\Leftrightarrow \pi.\sin x\neq \dfrac{\pi}{2}+k\pi$

$\Leftrightarrow \sin x\neq k+\dfrac{1}{2}$

$\tan (\pi.\sin x)=0$

$\Leftrightarrow \pi.\sin x=k\pi$

$\Leftrightarrow \sin x=k$ 

$\Rightarrow -1\le k\le 1\Rightarrow k\in \{ -1;0;1\}$

$k=-1\Rightarrow x=\dfrac{-\pi}{2}+k2\pi$

$k=0\Rightarrow x=k\pi$

$k=1\Rightarrow x=\dfrac{\pi}{2}+k2\pi$