Question

giải pt : 1) \(2sin^2x-sin3x+sin5x=0\)

             2)\(sin3x+sinx=2sin^2x\)

             3) \(cos5x+\sqrt{3}sin3x=cosx\)

Mong mn giúp mình ạ 

Answer 1

1.

\(2sin^2x+sin5x-sin3x=0\)

\(\Leftrightarrow2sin^2x+2cos4x.sinx=0\)

\(\Leftrightarrow2sinx\left(sinx+cos4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos4x=-sinx\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos4x=cos\left(\dfrac{\pi}{2}+x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\4x=\dfrac{\pi}{2}+x+k2\pi\\4x=-\dfrac{\pi}{2}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=-\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

Answer 2

2.

\(sin3x+sinx-2sin^2x=0\)

\(\Leftrightarrow2sin2x.cosx-2sin^2x=0\)

\(\Leftrightarrow4sinx.cos^2x-2sin^2x=0\)

\(\Leftrightarrow4sinx\left(1-sin^2x\right)-2sin^2x=0\)

\(\Leftrightarrow2sinx\left(2-2sin^2x-sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{-1+\sqrt{17}}{4}\\sinx=\dfrac{-1-\sqrt{17}}{4}< -1\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=arcsin\left(\dfrac{-1+\sqrt{17}}{4}\right)+k2\pi\\x=\pi-arcsin\left(\dfrac{-1+\sqrt{17}}{4}\right)+k2\pi\end{matrix}\right.\)

Answer 3

3.

\(cos5x-cosx+\sqrt{3}sin3x=0\)

\(\Leftrightarrow-2sin3x.sin2x+\sqrt{3}sin3x=0\)

\(\Leftrightarrow sin3x\left(-2sin2x+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\sin2x=\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=k\pi\\2x=\dfrac{\pi}{3}+k2\pi\\2x=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

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