Question

[Thanh Huyền Lê]

Help với ạ!!!!!

Answer 1

[Nguyễn Việt Lâm]

c.

ĐKXĐ: \(sinx\ne0\Rightarrow x\ne k\pi\)

\(1-\dfrac{\sqrt{3}cosx}{sinx}-4cosx=0\)

\(\Rightarrow sinx-\sqrt{3}cosx-4sinx.cosx=0\)

\(\Leftrightarrow sinx-\sqrt{3}cosx=2sin2x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin2x\)

\(\Leftrightarrow sin2x=sin\left(x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=x-\dfrac{\pi}{3}+k2\pi\\2x=\dfrac{4\pi}{3}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

Answer 2

[Nguyễn Việt Lâm]

a.

\(\Leftrightarrow3sin3x-4sin^34x-\sqrt{3}cos9x=2sin2x\)

\(\Leftrightarrow sin9x-\sqrt{3}cos9x=2sin2x\)

\(\Leftrightarrow\dfrac{1}{2}sin9x-\dfrac{\sqrt{3}}{2}cos9x=sin2x\)

\(\Leftrightarrow sin\left(9x-\dfrac{\pi}{3}\right)=sin2x\)

\(\Leftrightarrow\left[{}\begin{matrix}9x-\dfrac{\pi}{3}=2x+k2\pi\\9x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{21}+\dfrac{k2\pi}{7}\\x=\dfrac{4\pi}{33}+\dfrac{k2\pi}{11}\end{matrix}\right.\)