Bạn tham khảo pt 1 hộ mình nha. Chúc bạn học tốt~

Pt 1.

Bạn tham khảo phương trình 1 hộ mình nha. Chúc bạn học tốt

a) pt <=> - cos2x. tan²2x + 3.cos2x=0

      <=>  \(\dfrac{sin^22x}{-cos2x}\)+ 3cos2x =0

      <=>  sin²2x - 3cos²2x = 0

     <=> 1 - 4 cos²2x = 0

      <=> 1 - 4.\(\dfrac{1+cos4x}{2}\)= 0

      <=>  cos4x = \(\dfrac{-1}{2}\)

Chứng minh VT=VP cơ ạ

1: ĐKXĐ: 2x-1<>0

=>2x<>1

=>x<>1/2

=>TXĐ là D=R\{1/2}

2: ĐKXĐ: \(3x+\frac25\pi<>\frac{\pi}{2}+k\pi\)

=>\(3x<>\frac{\pi}{2}-\frac25\pi+k\pi=\frac{1}{10}\pi+k\pi\)

=>\(x<>\frac{1}{30}\pi+\frac{k\pi}{3}\)

=>TXĐ là D=R\{\(\frac{\pi}{30}+\frac{k\pi}{3}\) }

3: ĐKXĐ: \(2x-\frac13<>k\pi\)

=>\(2x<>\frac13+k\pi\)

=>\(x<>\frac16+\frac{k\pi}{2}\)

=>TXĐ là D=R\{\(\frac{k\pi}{2}+\frac16\) }

4: ĐKXĐ: sin x-cosx<>0

=>\(\sqrt2\cdot\sin\left(x-\frac{\pi}{4}\right)<>0\)

=>\(\sin\left(x-\frac{\pi}{4}\right)<>0\)

=>\(x-\frac{\pi}{4}<>k\pi\)

=>\(x<>\frac{\pi}{4}+k\pi\)

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

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

=>TXĐ là D=R\{\(\frac{k\pi}{2}\) }

6: ĐKXĐ: \(\begin{cases}1-\sin x\ge0\\ cosx<>0\end{cases}\Rightarrow\begin{cases}\sin x<=1\\ x<>\frac{\pi}{2}+k\pi\end{cases}\)

=>\(x<>\frac{\pi}{2}+k\pi\)

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

7: ĐKXĐ: \(\sin^2x-cos^2x<>0\)

=>\(cos^2x-\sin^2x<>0\)

=>cos2x<>0

=>\(2x<>\frac{\pi}{2}+k\pi\)

=>\(x<>\frac{\pi}{4}+\frac{k\pi}{2}\)

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

8: ĐKXĐ: \(\begin{cases}x<>\frac{\pi}{2}+k\pi\\ \sin x<>-1\end{cases}\Rightarrow\begin{cases}x<>\frac{\pi}{2}+k\pi\\ x<>-\frac{\pi}{2}+k2\pi\end{cases}\)

=>\(x<>\frac{\pi}{2}+k\pi\)

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

3.

\(4sinx.cosx-2sinx+1-2cosx=0\)

\(\Leftrightarrow2sinx\left(2cosx-1\right)-\left(2cosx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(2cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\cosx=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

4.

\(cosx-sinx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\-4sinx.cosx=2t^2-2\end{matrix}\right.\)

Pt trở thành: \(t+2t^2-2-1=0\Leftrightarrow2t^2+t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{3}{2}< -\sqrt{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2}cos\left(x+\frac{\pi}{4}\right)=-1\)

\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow...\)

5.

\(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=sinx\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sinx\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=x+k2\pi\\2x+\frac{\pi}{6}=\pi-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

6.

\(9sin^2x-5\left(1-sin^2x\right)-5sinx+4=0\)

\(\Leftrightarrow14sin^2x-5sinx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-\frac{1}{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=arcsin\left(-\frac{1}{7}\right)+k2\pi\\x=\pi-arcsin\left(-\frac{1}{7}\right)+k2\pi\end{matrix}\right.\)