Câu $1$. PT
$\Leftrightarrow \tan (\pi \sin x)=\tan (\frac{\pi}{2}-\pi \cos x)$
$\Leftrightarrow \pi \sin x=\frac{\pi}{2}-\pi \cos x + k\pi (k \in \mathbb{Z})$$\Leftrightarrow \sin x=\frac{1}{2}-\cos x + k (k \in \mathbb{Z})$ $\Leftrightarrow \sin x=\frac{1}{2}-\cos x + k (k \in \mathbb{Z})$ $\Leftrightarrow \sin x+\cos x=\frac{1}{2} + k (k \in \mathbb{Z})$ $\Leftrightarrow \sqrt 2 \sin \left ( x +\frac{\pi}{4} \right )=\frac{1}{2} + k (*) $ Ta biết rằng $\left| {\sqrt 2 \sin \left ( x +\frac{\pi}{4} \right )} \right| \le \sqrt 2 \forall x.$ Do đó $\left| {\frac{1}{2} + k} \right| \le \sqrt 2$. Mà $k \in \mathbb{Z} \implies k \in \left\{ {-1; 0} \right\}.$ + Với $k=0$. PT $(*)\Leftrightarrow \sin \left ( x +\frac{\pi}{4} \right )=\frac{1}{2\sqrt 2} \Leftrightarrow \left[ {\begin{matrix} x=\arcsin\frac{1}{2\sqrt 2} -\frac{\pi}{4} +k2\pi \\x=\frac{3\pi}{4}-\arcsin\frac{1}{2\sqrt 2} +k2\pi \end{matrix}} \right. (k \in \mathbb{Z})$+ Với $k=-1$. PT $(*)\Leftrightarrow \sin \left ( x +\frac{\pi}{4}
\right )=-\frac{1}{2\sqrt 2} \Leftrightarrow \left[ {\begin{matrix}
x=\arcsin\frac{-1}{2\sqrt 2} -\frac{\pi}{4} +k2\pi
\\x=\frac{3\pi}{4}-\arcsin\frac{-1}{2\sqrt 2} +k2\pi \end{matrix}}
\right. (k \in \mathbb{Z})$
Normal
0
false
false
false
EN-US
X-NONE
X-NONE
MicrosoftInternetExplorer4
Câu $1$. PT
$\Leftrightarrow \tan (\pi \sin x)=\tan (\frac{\pi}{2}-\pi \cos x)$
$\Leftrightarrow \pi \sin x=\frac{\pi}{2}-\pi \cos x + k\pi (k \in \mathbb{Z})$$\Leftrightarrow \sin x=\frac{1}{2}-\cos x + k (k \in \mathbb{Z})$ $\Leftrightarrow \sin x=\frac{1}{2}-\cos x + k (k \in \mathbb{Z})$ $\Leftrightarrow \sin x+\cos x=\frac{1}{2} + k (k \in \mathbb{Z})$ $\Leftrightarrow \sqrt 2 \sin \left ( x +\frac{\pi}{4} \right )=\frac{1}{2} + k (*) $ Ta biết rằng $\left| {\sqrt 2 \sin \left ( x +\frac{\pi}{4} \right )} \right| \le \sqrt 2 \forall x.$ Do đó $\left| {\frac{1}{2} + k} \right| \le \sqrt 2$. Mà $k \in \mathbb{Z} \implies k \in \left\{ {-1; 0} \right\}.$ + Với $k=0$. PT $(*)\Leftrightarrow \sin \left ( x +\frac{\pi}{4} \right )=\frac{1}{2\sqrt 2} \Leftrightarrow \left[ {\begin{matrix} x=\arcsin\frac{1}{2\sqrt 2} -\frac{\pi}{4} +k2\pi \\x=\frac{3\pi}{4}-\arcsin\frac{1}{2\sqrt 2} +k2\pi \end{matrix}} \right. (k \in \mathbb{Z})$+ Với $k=-1$. PT $(*)\Leftrightarrow \sin \left ( x +\frac{\pi}{4}
\right )=-\frac{1}{2\sqrt 2} \Leftrightarrow \left[ {\begin{matrix}
x=\arcsin\frac{-1}{2\sqrt 2} -\frac{\pi}{4} +k2\pi
\\x=\frac{3\pi}{4}-\arcsin\frac{-1}{2\sqrt 2} +k2\pi \end{matrix}}
\right. (k \in \mathbb{Z})$
Câu $1$. PT
$\Leftrightarrow \tan (\pi \sin x)=\tan (\frac{\pi}{2}-\pi \cos x)$
$\Leftrightarrow \pi \sin x=\frac{\pi}{2}-\pi \cos x + k\pi (k \in \mathbb{Z})$$\Leftrightarrow \sin x=\frac{1}{2}-\cos x + k (k \in \mathbb{Z})$ $\Leftrightarrow \sin x=\frac{1}{2}-\cos x + k (k \in \mathbb{Z})$ $\Leftrightarrow \sin x+\cos x=\frac{1}{2} + k (k \in \mathbb{Z})$ $\Leftrightarrow \sqrt 2 \sin \left ( x +\frac{\pi}{4} \right )=\frac{1}{2} + k (*) $ Ta biết rằng $\left| {\sqrt 2 \sin \left ( x +\frac{\pi}{4} \right )} \right| \le \sqrt 2 \forall x.$ Do đó $\left| {\frac{1}{2} + k} \right| \le \sqrt 2$. Mà $k \in \mathbb{Z} \implies k \in \left\{ {-1; 0} \right\}.$ + Với $k=0$. PT $(*)\Leftrightarrow \sin \left ( x +\frac{\pi}{4} \right )=\frac{1}{2\sqrt 2} \Leftrightarrow \left[ {\begin{matrix} x=\arcsin\frac{1}{2\sqrt 2} -\frac{\pi}{4} +k2\pi \\x=\frac{3\pi}{4}-\arcsin\frac{1}{2\sqrt 2} +k2\pi \end{matrix}} \right. (k \in \mathbb{Z})$+ Với $k=-1$. PT $(*)\Leftrightarrow \sin \left ( x +\frac{\pi}{4}
\right )=-\frac{1}{2\sqrt 2} \Leftrightarrow \left[ {\begin{matrix}
x=\arcsin\frac{-1}{2\sqrt 2} -\frac{\pi}{4} +k2\pi
\\x=\frac{3\pi}{4}-\arcsin\frac{-1}{2\sqrt 2} +k2\pi \end{matrix}}
\right. (k \in \mathbb{Z})$