Question

Solve for 0≤ x ≤ 180°: tan^2(x)-4sin^2(x)=0

Answer 1

\[\begin{array}{rcl}
\frac{\sin^2x}{\cos^2x}-4\sin^2x&=&0\\
\frac{\sin^2x-4\sin^2x\cos^2x}{\cos^2x}&=&0\\
\sin^2x-4\sin^2x\cos^2x&=&0\\
\sin^2x-4\sin^2x(1-\sin^2x)&=&0\\
-3\sin^2x+4\sin^4x&=&0\\
\sin^2x(4\sin^2x-3)&=&0\\
\end{array}\]\[\begin{array}{rclcrcl}
\sin^2x&=&0&\mbox{or}&4\sin^2x-3&=&0\\
\sin x&=&0&\mbox{or}&\sin x&=&\frac{\sqrt3}2\\
x=0&\mbox{or}&x=180^\circ&\mbox{or}&x=60^\circ&\mbox{or}&x=120^\circ
\end{array}\]

Answer 2

Thank you so much! Although there was a slightly different way to do it. Either way, Thanks!