Tập xác định của hàm số $y = \sin \sqrt{x}$ là:

Chọn B

$\sin \left(\pi x\right)=\cos \left(\frac{\pi }{3}+\pi x\right)$
$\iff \cos \left(\frac{\pi }{2}-\pi x\right)=\cos \left(\frac{\pi }{3}+\pi x\right)$ hay $\cos \left(\frac{\pi }{3}+\pi x\right)=\cos \left(\frac{\pi }{2}-\pi x\right)$
$\iff \left[\begin{array}{l}\frac{\pi }{3}+\pi x=\frac{\pi }{2}-\pi x+k2\pi \\ \frac{\pi }{3}+\pi x=-\frac{\pi }{2}+\pi x+k2\pi \left(l\right)\end{array}\right.$
$\iff 2\pi x=\frac{\pi }{6}+k2\pi$

$\iff x=\frac{1}{12}+k$

Chọn A

Ta có:$y=\frac{2\sin x+\cos x+1}{\sin x-2\cos x+3}$

$$\begin{gathered} {(\sin x - 2\cos x)}^2\le (1^2 +\hspace{0.17em}{2}^2) . ({\sin }^{2}x\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}{\cos }^{2}x)= 5 \\ \Rightarrow - \sqrt{5}\le \sin x - 2\cos x\le \sqrt{5} \\ \Rightarrow \sin x - 2\cos x \hspace{0.17em}+ 3 > 0\forall x \end{gathered}$$

+ $y=\frac{2\sin x+\cos x+1}{\sin x-2\cos x+3}$

$\iff$y. (sinx - 2cosx + 3) = 2sinx + cosx + 1 

$\iff$(y-2)  . sinx - (2y+ 1)cosx = 1 - 3y  (*)

Để phương trình (*) có nghiệm điều kiện là:

$$\begin{gathered} {( y- 2)}^2+\hspace{0.17em}{(2y +\hspace{0.17em}1)}^2\ge {(\hspace{0.17em}1- 3y)}^2\iff y^2- 4y +\hspace{0.17em}4 +\hspace{0.17em}4y^2+ 4y +\hspace{0.17em}1\ge 1- 6y +\hspace{0.17em}9y^2 \\ \iff - 4y^2+\hspace{0.17em}6y +\hspace{0.17em}4 \ge 0 \\ \iff \frac{-1}{2}\le y\le 2 \end{gathered}$$