Question

Lm hộ mks với 😭😭

Answer 1

\(4sin^2x-2\left(\sqrt{3}+\sqrt{2}\right)sinx+\sqrt{6}=0\)

\(\Leftrightarrow4sin^2x-2\sqrt{3}sinx-2\sqrt{2}sinx+\sqrt{6}=0\)

\(\Leftrightarrow2sinx\left(2sinx-\sqrt{3}\right)-\sqrt{2}\left(2sinx-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left(2sinx-\sqrt{3}\right)\left(2sinx-\sqrt{2}\right)=0\)

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

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