ĐKXĐ: \(cosx\ne0\)

Đặt \(sinx=t\)

\(\Rightarrow5t-2=3\left(1-t\right).\frac{t^2}{1-t^2}\)

\(\Leftrightarrow5t-2=\frac{3t^2}{1+t}\)

\(\Leftrightarrow\left(5t-2\right)\left(1+t\right)=3t^2\)

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

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

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

ĐKXĐ: sin x<>0

=>x<>kΩ

Đặt \(t=\sin x+\frac{1}{\sin x}\)

=>\(\left|t\right|\ge2\cdot\sqrt{\sin x\cdot\frac{1}{\sin x}}=2\)

\(t^2=\left(\sin x+\frac{1}{\sin x}\right)^2=\sin^2x+\frac{1}{\sin^2 x}+2\)

\(\Rightarrow\sin^2x+\frac{1}{\sin^2x}=t^2-2\)

Phương trình trở thành: \((t^2-2)+4t-7=0\)

=>\(t^2+4t-9=0\)

=>\(t^2+4t+4=13\)

=>\(\left(t+2\right)^2=13\)

=>\(\left[\begin{array}{l}t+2=\sqrt{13}\\ t+2=-\sqrt{13}\end{array}\right.\Rightarrow\left[\begin{array}{l}t=-2+\sqrt{13}\left(loại\right)\\ t=-2-\sqrt{13}\left(nhận\right)\end{array}\right.\)

\(t=-2-\sqrt{13}\)

=>\(\sin x + \frac{1}{\sin x} = -2 - \sqrt{13}\)

=>\(\sin^2 x + (2 + \sqrt{13})\sin x + 1 = 0\) (1)

\(\Delta = (2 + \sqrt{13})^2 - 4 = 4 + 4\sqrt{13} + 13 - 4 = 13 + 4\sqrt{13}\)

Do đó: Phương trình (1)có hai nghiệm phân biệt là:

\(\left[\begin{array}{l} \sin x = \frac{-(2+\sqrt{13}) + \sqrt{13+4\sqrt{13}}}{2} \approx -0,192 \quad (\text{Thỏa mãn } -1 \le \sin x \le 1) \\ \sin x = \frac{-(2+\sqrt{13}) - \sqrt{13+4\sqrt{13}}}{2} \approx -5,413 \quad (\text{Loại do } \sin x < -1) \end{array}\right.\)

=>\(\left[\begin{array}{l} x = \arcsin\left(\frac{-(2+\sqrt{13}) + \sqrt{13+4\sqrt{13}}}{2}\right) + k2\pi \\ x = \pi - \arcsin\left(\frac{-(2+\sqrt{13}) + \sqrt{13+4\sqrt{13}}}{2}\right) + k2\pi \end{array}\right. \quad (k \in \mathbb{Z})\)

Ta có : \(2cos^2x+2\sqrt{3}sinx.cosx+1=3\left(sinx+\sqrt{3}cosx\right)\) 

\(\Leftrightarrow3cos^2x+sin^2x+2\sqrt{3}sinxcosx=3\left(sinx+\sqrt{3}cosx\right)\) 

\(\Leftrightarrow\left(\sqrt{3}cosx+sinx\right)^2=3\left(\sqrt{3}cosx+sinx\right)\) 

\(\Leftrightarrow\left(\sqrt{3}cosx+sinx\right)\left(\sqrt{3}cosx+sinx-3\right)=0\) 

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3}cosx+sinx=0\\\sqrt{3}cos+sinx=3\end{matrix}\right.\) 

Thấy : \(-1\le sinx;cosx\le1\Rightarrow\sqrt{3}cosx+sinx\le1+\sqrt{3}< 3\) 

Do đó : \(\sqrt{3}cosx+sinx=0\)  \(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx=0\)

\(\Leftrightarrow sin\dfrac{\pi}{3}.cosx+cos\dfrac{\pi}{3}sinx=0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\Leftrightarrow x=\dfrac{-\pi}{3}+k\pi\) ( k thuộc Z ) 

Vậy ... 

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 = \(\dfrac{-1}{2}\)

\(\sqrt{2}\left(2cos^2x-3sin2x\right)=4cosx.sin2x+2\left(sinx-cosx\right)\)

\(\Leftrightarrow\left(2\sqrt{2}cos^2x+2cosx\right)-3\sqrt{2}sin2x-4cosx.sin2x-2sinx=0\)

\(\Leftrightarrow2cosx\left(\sqrt{2}cosx+1\right)-6\sqrt{2}sinx.cosx-4cosx^2.sinx-2sinx=0\)

\(\Leftrightarrow2cosx\left(\sqrt{2}cosx+1\right)-2sinx\left(4cos^2x+3\sqrt{2}cosx+1\right)=0\)

\(\Leftrightarrow2cosx\left(\sqrt{2}cosx+1\right)-2sinx\left(\sqrt{2}cosx+1\right)\left(2\sqrt{2}cosx+1\right)=0\)

\(\Leftrightarrow\left(2cosx-4\sqrt{2}cosx.sinx-2sinx\right)\left(\sqrt{2}cosx+1\right)=0\)

\(\Leftrightarrow\left[2\sqrt{2}-2\sqrt{2}\left(cosx-sinx\right)^2+2\left(cosx-sinx\right)\right]\left(\sqrt{2}cosx+1\right)=0\)

Đặt \(t=cosx-sinx\left(t\in\left[-\sqrt{2};\sqrt{2}\right]\right)\)

\(pt\Leftrightarrow\left[{}\begin{matrix}cosx=-\dfrac{1}{\sqrt{2}}\\\sqrt{2}t^2-t-\sqrt{2}=0\end{matrix}\right.\)

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