Đáp án B

Vậy PT có 1 nghiệm thuộc (0; π )

Đáp án B

ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\sin^2x\ne1\end{matrix}\right.\) \(\Leftrightarrow cosx\ne0\)

\(\Leftrightarrow x\ne\frac{\pi}{2}+k\pi\)

TXĐ là D=R

Khi x∈D thì -x∈D

\(f\left(-x\right)=\left(m+1\right)\cdot cos\left(-x\right)+\tan\left(-x\right)-3\cdot\left(-x\right)+\left(m^2-1\right)\cdot\left(-x\right)^2\)

\(=\left(m+1\right)\cdot cosx-tanx+3x+\left(m^2-1\right)\cdot x^2\)

Để đây là hàm số lẻ thì f(x)=-f(x)

=>\(\left(m+1\right)\cdot cosx+\tan x-3x+\left(m^2-1\right)\cdot x^2=-\left(m+1\right)\cdot cosx+\tan x-3x-\left(m^2-1\right)\cdot x^2\)

=>2(m+1)*cosx+2(m^2-1)*x^2=0

=>(m+1)[2*cosx+2x^2(m-1)]=0

=>m+1=0

=>m=-1

ĐKXĐ: \(\left\{{}\begin{matrix}\frac{1+tanx}{1-tanx}\ge0\\tanx\ne1\\cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-1\le tanx< 1\\x\ne\frac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Rightarrow-\frac{\pi}{4}+k\pi\le x< \frac{\pi}{4}+k\pi\)

d/

ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow\frac{sin\left(3x-x\right)}{cos^2x}=2\sqrt{3}\)

\(\Leftrightarrow\frac{sin2x}{cos^2x}=2\sqrt{3}\)

\(\Leftrightarrow\frac{2sinx.cosx}{cos^2x}=2\sqrt{3}\)

\(\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\)

\(\Leftrightarrow tanx=\sqrt{3}\)

\(\Rightarrow x=\frac{\pi}{3}+k\pi\)

c/

ĐKXĐ: \(sin2x\ne0\)

\(\Leftrightarrow\frac{\frac{sinx}{cosx}-sinx}{sin^3x}=\frac{1}{cosx}\)

\(\Leftrightarrow sinx-sinx.cosx=sin^3x\)

\(\Leftrightarrow1-cosx=sin^2x\)

\(\Leftrightarrow1-cosx=1-cos^2x\)

\(\Leftrightarrow cos^2x-cosx=0\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\)