Question

giải các pt

a) \(\left(2sin^2x-1\right)tan^22x+3\left(2cos^2x-1\right)=0\)

b) \(tanx+tan2x=\frac{2sin3x}{sin2x}\)

c) \(1+sinx.cos2x=sinx+cos2x\)

d) \(tanx=1-cos2x\)

Answer 1

a/

DKXD: ...

\(\Leftrightarrow-cos2x.tan^22x+3.cos2x=0\)

\(\Leftrightarrow cos2x\left(3-tan^22x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\tan^22x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\tan2x=\sqrt{3}\\tan2x=-\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\2x=\frac{\pi}{3}+k\pi\\2x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\pm\frac{\pi}{6}+\frac{k\pi}{2}\end{matrix}\right.\)

Answer 2

b/

DKXD: ...

\(\Leftrightarrow\frac{sinx}{cosx}+\frac{sin2x}{cos2x}-\frac{2sin3x}{sin2x}=0\)

\(\Leftrightarrow\frac{sinx.cos2x+sin2x.cosx}{cosx.cos2x}-\frac{2sin3x}{sin2x}=0\)

\(\Leftrightarrow\frac{sin\left(2x+x\right)}{cosx.cos2x}-\frac{2sin3x}{sin2x}=0\)

\(\Leftrightarrow\frac{sin3x}{cosx.cos2x}-\frac{2sin3x}{sin2x}=0\)

\(\Leftrightarrow sin3x\left(\frac{1}{cosx.cos2x}-\frac{2}{sin2x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\left(1\right)\\2cosx.cos2x=sin2x\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow3sinx-4sin^3x=0\) (tìm nghiệm thẳng bằng \(3x=k\pi\) rồi dựa vào đường tròn lượng giác loại nghiệm cũng được)

\(\Leftrightarrow sinx\left(3-4sin^2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(l\right)\\sinx=\pm\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{3}+k2\pi\\x=\frac{2\pi}{3}+k2\pi\\x=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow2cosx.cos2x=2sinx.cosx\)

\(\Leftrightarrow2cosx\left(cos2x-sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(l\right)\\cos2x=sinx=cos\left(\frac{\pi}{2}-x\right)\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=-\frac{\pi}{2}+k2\pi\left(l\right)\end{matrix}\right.\)

Answer 3

c/

\(\Leftrightarrow sinx.cos2x-sinx+1-cos2x=0\)

\(\Leftrightarrow sinx\left(cos2x-1\right)-\left(cos2x-1\right)=0\)

\(\Leftrightarrow\left(sinx-1\right)\left(cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\cos2x=1\end{matrix}\right.\)

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

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

Answer 4

d/

ĐKXĐ: ...

\(\Leftrightarrow tanx-1+cos2x=0\)

\(\Leftrightarrow\frac{sinx}{cosx}-1-\left(sin^2x-cos^2x\right)=0\)

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

\(\Leftrightarrow\left(sinx-cosx\right)\left(\frac{1}{cosx}-sinx-cosx\right)=0\)

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

\(\left(1\right)\Rightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\)

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

\(\left(2\right)\Leftrightarrow1-sinx.cosx-cos^2x=0\)

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

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

\(\Leftrightarrow sinx=0\Rightarrow x=k\pi\)