Đáp án C

Đáp án C

\(\frac{2}{sin4x}-tan2x=\frac{2}{2sin2x.cos2x}-\frac{sin2x}{cos2x}=\frac{1}{cos2x}\left(\frac{1}{sin2x}-sin2x\right)\)

\(=\frac{1}{cos2x}\left(\frac{1-sin^22x}{sin2x}\right)=\frac{1}{cos2x}\frac{cos^22x}{sin2x}=\frac{cos2x}{sin2x}=cot2x\)

\(A=\frac{1}{2}\left(\frac{sin^2x}{cos^2x}-1\right)\frac{cosx}{sinx}+cos4x.cot2x+sin4x\)

\(A=\frac{-1}{2}\left(\frac{cos^2x-sin^2x}{cos^2x}\right)\frac{cosx}{sinx}+cos4x.cot2x+sin4x\)

\(A=\frac{-cos2x}{2cosx.sinx}+cos4x.cot2x+sin4x\)

\(A=-cot2x+cos4x.cot2x+sin4x\)

\(A=cot2x\left(cos4x-1\right)+sin4x\)

\(A=\frac{cos2x}{sin2x}.\left(1-2sin^22x-1\right)+sin4x\)

\(A=\frac{-2cos2x.sin^22x}{sin2x}+sin4x\)

\(A=-sin4x+sin4x=0\)

     \(sin^3x+cos^3x=\dfrac{\sqrt{2}}{2}\)   (_*)

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

\(\Rightarrow\) \(\left(sinx+cosx\right)\left(1-sinx\cdot cosx\right)=\dfrac{\sqrt{2}}{2}\)    (1)

 Đặt \(t=sinx+cosx\left(t\le\left|\sqrt{2}\right|\right)\)

       \(\Rightarrow\)\(sinx\cdot cosx=\dfrac{t^2-1}{2}\)

 Khi đó (1) thành: \(t\cdot\dfrac{1-t^2}{2}=\dfrac{\sqrt{2}}{2}\) \(\Rightarrow t=-\sqrt{2}\left(tm\right)\)

               \(\Rightarrow sinx+cosx=-\sqrt{2}\)

              \(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-\sqrt{2}\)

              \(\Rightarrow sin\left(x+\dfrac{\pi}{4}\right)=-1\)

             \(\Rightarrow x+\dfrac{\pi}{4}=-\dfrac{\pi}{2}+k2\pi\) \(\Rightarrow x=-\dfrac{3\pi}{4}+k2\pi\left(k\in Z\right)\)

Ta biến đổi :

\(f\left(x\right)=\frac{\sin3x\sin4x}{\tan x+\cot2x}=\frac{\sin3x\sin4x}{\frac{\sin x.\sin2x+\cos x.\cos2x}{\cos x.\sin2x}}=\frac{\sin3x\sin4x}{\frac{\cos x}{\cos x.\sin2x}}=\sin3x\sin4x\sin2x\)

\(=\frac{1}{2}\left(\cos x-\cos7x\right)\sin2x=\frac{1}{2}\left[\sin2x\cos x-\cos7x\sin2x\right]=\frac{1}{4}\left(\sin3x+\sin x-\sin9x+\sin5x\right)\)

Do đó :

\(I=\int\left(\frac{1}{4}\left(\sin3x+\sin x-\sin9x+\sin5x\right)\right)dx=-\frac{1}{2}\cos3x-\frac{1}{4}\cos x+\frac{1}{9}\cos9x-\frac{1}{5}\cos5x+C\)

D.time

Câu 1:

ĐKXĐ: \(sin4x\ne0\Rightarrow x\ne\frac{k\pi}{4}\)

\(\Leftrightarrow\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=2\Leftrightarrow4sinx.cos2x+2cos2x=2\)

\(\Leftrightarrow cos2x\left(2sinx+1\right)=1\Leftrightarrow\left(1-2sin^2x\right)\cdot\left(2sinx+1\right)=1\)

\(\Leftrightarrow2sinx-4sin^3x-2sin^2x=0\)

\(\Leftrightarrow sinx\left(-2sin^2x-sinx+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=0\left(l\right)\\sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

b/

ĐKXĐ: \(sin2x\ne0\Rightarrow x\ne\frac{k\pi}{2}\)

\(\Leftrightarrow\frac{sin^4x+cos^4x}{5}=\frac{1}{2}cos2x-\frac{1}{8}\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2\left(sinx.cosx\right)^2=\frac{5}{2}cos2x-\frac{5}{8}\)

\(\Leftrightarrow1-\frac{1}{2}sin^22x=\frac{5}{2}cos2x-\frac{5}{8}\)

\(\Leftrightarrow1-\frac{1}{2}\left(1-cos^22x\right)=\frac{5}{2}cos2x-\frac{5}{8}\)

\(\Leftrightarrow\frac{1}{2}cos^22x-\frac{5}{2}cos2x+\frac{9}{8}=0\)

\(\Rightarrow\left[{}\begin{matrix}cos2x=\frac{9}{2}>1\left(l\right)\\cos2x=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow x=\pm\frac{\pi}{6}+k\pi\)