Question

\(sin^4x-cos^2x=1\\ \dfrac{3}{cos^2x}+2\sqrt{3}.tgx-6=0\)

Answer 1

a: \(\Leftrightarrow1-cos^4x-cos^2x=1\)

\(\Leftrightarrow cos^2x\left(cos^2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\\cosx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\x=k2\Pi\\x=\Pi+k2\Pi\end{matrix}\right.\)

b: \(\Leftrightarrow3\left(1+\tan^2x\right)+2\sqrt{3}tanx-6=0\)

\(\Leftrightarrow3\cdot tan^2x+2\sqrt{3}\cdot tanx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\dfrac{\sqrt{3}}{3}\\tanx=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{6}+k\Pi\\x=-\dfrac{\Pi}{3}+k\Pi\end{matrix}\right.\)