\(\sqrt{3}cos5x-2sin3x.cos2x-sinx=0\)
⇔ \(\sqrt{3}cos5x-\left(sinx+sin5x\right)-sinx=0\)
⇔ \(\sqrt{3}cos5x-sin5x-2sinx=0\)
⇔ \(2sin\left(\dfrac{\pi}{3}-5x\right)=2sinx\)
⇔ \(2sin\left(5x-\dfrac{\pi}{3}\right)=2sin\left(-x\right)\)
Giải nốt nhé
\(\sqrt{3}\)Cos5x - 2Sin3x.Cos2x - Sinx =0
\(\Leftrightarrow\sqrt{3}\)Cos5x -(Sin5x +Sinx ) -Sinx =0
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}\)Cos5x - \(\dfrac{1}{2}\)Sin5x =Sinx
\(\Leftrightarrow\)Sin (\(\dfrac{\pi}{3}\) - 5x )=Sinx
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\pi}{3}-5x=x+k2\pi\\\dfrac{\pi}{3}-5x=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\pi}{18}-k\dfrac{\pi}{3}\\x=\dfrac{-\pi}{6}-k\dfrac{\pi}{2}\end{matrix}\right.\)
\(\Rightarrow\) Vậy pt có 2 ng x=\(\dfrac{\pi}{18}-k\dfrac{\pi}{3}\)và x=\(\dfrac{-\pi}{6}-k\dfrac{\pi}{2}\),k\(\varepsilon\)Z
