Câu hỏi:

y = tan2x xác định \(\Leftrightarrow 2x\ne \frac{\pi }{2}+k\pi \Leftrightarrow x\ne \frac{\pi }{4}+\frac{k\pi }{2}\,\,\left( k\in Z \right).\)

Chọn C.

TXĐ: \(x\ne \frac{\pi }{2}+k\pi \,\,\left( k\in Z \right)\Rightarrow D=R\backslash \left\{ \frac{\pi }{2}+k\pi \,\,\left( k\in Z \right) \right\}\)

\(\forall x\in D\Rightarrow -x\in D\) ta có:

\(\begin{array}{l}f\left( x \right) = \sin x + 2\tan x\\f\left( { - x} \right) = \sin \left( { - x} \right)+ 2\tan \left( { - x} \right) =  - \sin x - 2\tan x =  - f\left( x \right).\end{array}\)

Vậy hàm số là hàm lẻ.

Chọn B.

Ta có: \(y=4\sin x-3\cos x=5\left( \frac{4}{5}\sin x-\frac{3}{5}\cos x \right)\)

Đặt \(\cos \alpha =\frac{4}{5};\sin \alpha =\frac{3}{5}\) ta có: \(y=5\left( \sin x\cos \alpha -\cos x\sin \alpha  \right)=5\sin \left( x-\alpha  \right).\)

Ta có: \(-1\le \sin \left( x-\alpha  \right)\le 1\Leftrightarrow -5\le 5\sin \left( x-\alpha  \right)\le 5\) .

Vậy \(M=5,m=-5.\)

Chọn A.

\(\cos x=\frac{1}{2}=\cos \frac{\pi }{3}\Leftrightarrow x=\pm \frac{\pi }{3}+k2\pi \,\,\left( k\in Z \right).\)

Chọn D.

Ta có: \(\sin x = \frac{2}{3} \Leftrightarrow \left[ \begin{array}{l}x = \arcsin \frac{2}{3} + k2\pi \\x = \pi  - \arcsin \frac{2}{3} + k2\pi \end{array} \right.\,\,\left( {k \in Z} \right)\)

\(\begin{array}{l}

- \pi < \arcsin \frac{2}{3} + k2\pi < \pi \mathop \Leftrightarrow \limits^{k \in Z} \frac{{ - \pi - \arcsin \frac{2}{3}}}{{2\pi }} < k < \frac{{\pi - \arcsin \frac{2}{3}}}{{2\pi }}\mathop \Leftrightarrow \limits^{k \in Z} - 0,61 < k < 0,38 \Leftrightarrow k = 0\\

\Rightarrow x = \arcsin \frac{2}{3} = \arcsin \frac{2}{3}.\\

- \pi < \pi - \arcsin \frac{2}{3} + k2\pi < \pi \mathop \Leftrightarrow \limits^{k \in Z} \frac{{ - 2\pi + \arcsin \frac{2}{3}}}{{2\pi }} < k < \frac{{\arcsin \frac{2}{3}}}{{2\pi }}\mathop \Leftrightarrow \limits^{k \in Z} - 0,88 < k < 0,12\mathop \Leftrightarrow \limits^{k \in Z} k = 0.\\

\Rightarrow x = \pi - \arcsin \frac{2}{3} = \pi - \arcsin \frac{2}{3}.

\end{array}\)

 Vậy có 2 nghiệm thỏa mãn yêu cầu bài toán.

Chọn C.