Question

[Cao Văn Đạt]

Giải giúp e mấy vd vs

Answer 1

[Lê Thị Thục Hiền]

VD15:

Pt \(\Leftrightarrow2\left(1-sin^2x\right)=sinx+1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=\dfrac{-\pi}{2}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy...

CD16:

Pt \(\Leftrightarrow\dfrac{1-cos6x}{2}-\dfrac{1-cos4x}{2}-\dfrac{1-cos2x}{2}=0\)

\(\Leftrightarrow-1-cos6x+cos4x+cos2x=0\)

\(\Leftrightarrow-\left(1+cos6x\right)+\left(cos4x+cos2x\right)=0\)

\(\Leftrightarrow-2.cos^23x+2.cos3x.cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cos3x=cosx\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy...

VD17:

Pt\(\Leftrightarrow2.sin3x.cosx=2.cosx\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(\(k\in Z\))

Vậy...