Question

tan^2(th)=√3tan(th)

Solve this equation on 0<=th<2pi

th stands for theta

Answer 1

\(\bf tan^2(\theta)=\sqrt{3}\ tan(\theta)\implies \cfrac{tan^2(\theta)}{tan(\theta)}=\sqrt{3}\implies \cfrac{\cancel{ tan(\theta) }tan(\theta)}{\cancel{ tan(\theta) }}=\sqrt{3}\)

then take \(\bf tan^{-1}\) to both sides.. see what you get

Answer 2

keeping in mind that \(\begin{array}{cccl}

Function&{\color{brown}{ Domain}}&{\color{blue}{ Range}}\\
\hline\\
{\color{blue}{ y}}=sin^{-1}({\color{brown}{ \theta}})&-1\le {\color{brown}{ \theta}} \le 1&-\frac{\pi}{2}\le {\color{blue}{ y}}\le \frac{\pi}{2}
\\ \quad \\
{\color{blue}{ y}}=cos^{-1}({\color{brown}{ \theta}})&-1\le {\color{brown}{ \theta}} \le 1& 0 \le {\color{blue}{ y}}\le \pi
\\ \quad \\
{\color{blue}{ y}}=tan^{-1}({\color{brown}{ \theta}})&-\infty\le {\color{brown}{ \theta}} \le +\infty &-\frac{\pi}{2}\le {\color{blue}{ y}}\le \frac{\pi}{2}
\end{array}\)

Answer 3

I'm not sure what excatly am i taking the tan^(-1) of?