Question

((3)^1/2)*tan(theta)=2sin(theta) solve the equation for the domain 0< or equal to (theta)< or equal to 2Pi

Answer 1

\[\sqrt{3} \tan \theta=2\sin \theta, 0\le \theta \le360 \]

Answer 2

\[
\sqrt{3} \tan (\theta )=2 \sin
(\theta )\\
\frac{\sqrt{3} \sin (\theta
)}{\cos (\theta )}-2 \sin
(\theta )=0\\
\sin (\theta )
\left(\frac{\sqrt{3}}{\cos
(\theta )}-2\right)=0\\
\implies
\sin (\theta )=0 \\
or\\
\frac{\sqrt{3}}{\cos (\theta
)}=2\\
\cos (\theta
)=\frac{\sqrt{3}}{2}

\]
The rest should be easy

Answer 3

The above gives rise to
\[
\theta = 0\\
\theta = \pi\\
\theta = 2\pi\\
\theta = \frac \pi 6\\
\theta =- \frac \pi 6+ 2 \pi =\frac {11 \pi}6


\]

Answer 4

why is theta equal to 0 and 2pi

Answer 5

The sin is equal zero at 0, pi and 2 pi

Answer 6

The roots I stated above are all between 0 and 2 pi included

Answer 7

oh ok thank you :D