hộ vs ae ơi

a: ĐKXĐ: sin x-cos3x<>0

=>sin x<>cos3x

=>\(cos3x<>cos\left(\frac{\pi}{2}-x\right)\)

=>\(\begin{cases}3x<>\frac{\pi}{2}-x+k2\pi\\ 3x<>-\frac{\pi}{2}+x+k2\pi\end{cases}\Rightarrow\begin{cases}4x<>\frac{\pi}{2}+k2\pi\\ 2x<>-\frac{\pi}{2}+k2\pi\end{cases}\)

=>\(\begin{cases}x<>\frac{\pi}{8}+\frac{k\pi}{2}\\ x<>-\frac{\pi}{4}+k\pi\end{cases}\)

=>TXĐ là D=R\{\(\frac{\pi}{8}+\frac{k\pi}{2};-\frac{\pi}{4}+k\pi\) }

b: ĐKXĐ: \(\begin{cases}x<>\frac{\pi}{2}+k\pi\\ cos5x+cosx<>0\end{cases}\)

=>\(\begin{cases}x<>\frac{\pi}{2}+k\pi\\ cos5x<>-cosx=cos\left(\pi-x\right)\end{cases}\)

=>\(\begin{cases}x<>\frac{\pi}{2}+k\pi\\ 5x<>\pi-x+k2\pi\\ 5x<>x-\pi+k2\pi\end{cases}\Rightarrow\begin{cases}x<>\frac{\pi}{2}+k\pi\\ 6x<>\pi+k2\pi\\ 4x<>-\pi+k2\pi\end{cases}\)

=>\(\begin{cases}x<>\frac{\pi}{2}+k\pi\\ x<>\frac{\pi}{6}+\frac{k\pi}{3}\\ x<>-\frac{\pi}{4}+\frac{k\pi}{2}\end{cases}\)

=>TXĐ là D=R\{\(\frac{\pi}{2}+k\pi;\frac{\pi}{6}+\frac{k\pi}{3};-\frac{\pi}{4}+\frac{k\pi}{2}\) }

a: \(VT=\dfrac{cot^2x}{1+cot^2x}\cdot\dfrac{1+tan^2x}{tan^2x}\)

\(=\dfrac{cot^2x}{\dfrac{1}{sin^2x}}\cdot\dfrac{\dfrac{1}{cos^2x}}{tan^2x}\)

\(=\dfrac{cot^2x}{tan^2x}\cdot\dfrac{1}{cos^2x}:\dfrac{1}{sin^2x}\)

\(=\dfrac{cot^2x}{tan^2x}\cdot\dfrac{sin^2x}{cos^2x}\)

\(=cot^2x\)

\(VP=\dfrac{tan^2x+cot^2x}{1+tan^4x}=\dfrac{\dfrac{sin^2x}{cos^2x}+\dfrac{cos^2x}{sin^2x}}{1+\dfrac{sin^4x}{cos^4x}}\)

\(=\dfrac{sin^4x+cos^4x}{sin^2x\cdot cos^2x}:\dfrac{cos^4x+sin^4x}{cos^4x}\)

\(=\dfrac{sin^4x+cos^4x}{sin^2x\cdot cos^2x}\cdot\dfrac{cos^4x}{cos^4x+sin^4x}=\dfrac{cos^2x}{sin^2x}=cot^2x\)

=>VT=VP

b:

\(\dfrac{tan^2x-cos^2x}{sin^2x}+\dfrac{cot^2x-sin^2x}{cos^2x}\)

\(=\dfrac{\left(\dfrac{sinx}{cosx}\right)^2-cos^2x}{sin^2x}+\dfrac{\left(\dfrac{cosx}{sinx}\right)^2-sin^2x}{cos^2x}\)

\(=\dfrac{sin^2x-cos^4x}{cos^2x\cdot sin^2x}+\dfrac{cos^2x-sin^4x}{sin^2x\cdot cos^2x}\)

\(=\dfrac{sin^2x+cos^2x-cos^4x-sin^4x}{cos^2x\cdot sin^2x}\)

\(=\dfrac{1-\left(cos^2x+sin^2x\right)^2+2\cdot cos^2x\cdot sin^2x}{cos^2x\cdot sin^2x}\)

\(=\dfrac{2\cdot cos^2x\cdot sin^2x}{cos^2x\cdot sin^2x}=2\)

\(\frac{1}{\sin a}=\frac{1}{\sin\left(2\cdot\frac{a}{2}\right)}=\cot\left(\frac{a}{2}\right)-\cot a\)

\(\frac{1}{\sin2a}=\cot a-\cot2a\)

\(\frac{1}{\sin4a}=\cot2a-\cot4a\)

\(\frac{1}{\sin8a}=\cot4a-\cot8a\)

Ta có: \(S=\frac{1}{\sin a}+\frac{1}{\sin2a}+\frac{1}{\sin4a}+\frac{1}{\sin8a}\)

\(=\cot\left(\frac{a}{2}\right)-\cot a+\cot a-\cot2a+\cot2a-\cot4a+\cot4a-\cot8a\)

\(=\cot\left(\frac{a}{2}\right)-\cot8a\)

\(\dfrac{1+cos2x-sin2x}{cos2x}=\dfrac{1+2cos^2x-1-2sinx.cosx}{cos^2x-sin^2x}=\dfrac{2cosx\left(cosx-sinx\right)}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)

\(=\dfrac{2cosx}{cosx+sinx}=\dfrac{2}{\dfrac{cosx}{cosx}+\dfrac{sinx}{cosx}}=\dfrac{2}{1+tanx}\)

b/ ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow\left(1-\frac{sinx}{cosx}\right)\left(1+sinx\right)=1+\frac{sinx}{cosx}\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(1+sinx\right)=sinx+cosx\)

\(\Leftrightarrow cosx+sinx.cosx-sinx-sin^2x=sinx+cosx\)

\(\Leftrightarrow sin^2x+2sinx-sinx.cosx=0\)

\(\Leftrightarrow sinx\left(sinx-cosx+2\right)=0\)

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

Xét \(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=-2\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=-\sqrt{2}< -1\) (vô nghiệm)

a/ ĐKXĐ: \(sin4x\ne0\)

\(\frac{sinx}{cosx}+\frac{cos2x}{sin2x}=\frac{2cos4x}{sin4x}\)

\(\Leftrightarrow2sin^2x.cos2x+2cos^22x=2cos4x\)

\(\Leftrightarrow\left(1-cos2x\right)cos2x+2cos^22x=4cos^22x-2\)

\(\Leftrightarrow3cos^22x-cos2x-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\left(l\right)\\cos2x=-\frac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow2x=\pm arccos\left(-\frac{2}{3}\right)+k2\pi\)

\(\Leftrightarrow x=\pm\frac{1}{2}arccos\left(-\frac{2}{3}\right)+k\pi\)

Chọn A.

Ta có C = (1-sin²x) cot²x + 1 - cot²x.

= (1 - sin²x - 1)  cot²x + 1

= -sin²x.cot²x + 1 = -cos²x + 1 = sin²x.

    1 + sinx + cosx + sin2x + cos2x = 0

<=> sin^2x+ cos^2 x + ( sinx+cosx) + 2.sinx.cosx + ( cos^2 x - sin^2 x)=0

<=> 2 cos^2 x + 2sinx.cosx + sinx + cosx =0

<=> 2cosx ( cos x + sinx) + sinx + cosx = 0

<=> ( cosx + sinx ) (2 cos x + 1 ) = 0

<=> cosx + sinx = 0 hoặc 2cosx + 1 =0