Question

[Hạ Băng Băng]

Giải giúp mình với ạ

Answer 1

[Hồng Phúc]

1.

\(sin^3x-\sqrt{3}cos^3x=sinx.cos^2x-\sqrt{3}sin^2x.cosx\)

\(\Leftrightarrow sin^3x+\sqrt{3}sin^2x.cosx-\sqrt{3}cos^3x-sinx.cos^2x=0\)

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

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

\(\Leftrightarrow\left(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\right).\left(cos^2x-sin^2x\right)=0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right).cos2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)=0\\cos2x=0\end{matrix}\right.\)

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

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

Answer 2

[Hồng Phúc]

2.

\(sin3x-\sqrt{3}cos3x+2=4cos^2x\)

\(\Leftrightarrow4cos^2x-2+\sqrt{3}cos3x-sin3x=0\)

\(\Leftrightarrow2cos^2x-1+\dfrac{\sqrt{3}}{2}cos3x-\dfrac{1}{2}sin3x=0\)

\(\Leftrightarrow cos2x+cos\left(3x+\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow2cos\left(\dfrac{5x}{2}+\dfrac{\pi}{12}\right).cos\left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\dfrac{5x}{2}+\dfrac{\pi}{12}\right)=0\\cos\left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5x}{2}+\dfrac{\pi}{12}=\dfrac{\pi}{2}+k\pi\\\dfrac{x}{2}+\dfrac{\pi}{12}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

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