Question

Solve each equation over the interval [0, 2pi) tan^2x-3=0

Answer 1

\[
\tan^2(x)= 3\\
\tan (x)= \pm \sqrt3\\
\]
Can you finish it?

Answer 2

Hint
\[

\tan(\pi/3)=\sqrt 3
\]

Answer 3

and also
\[
\tan(-\pi/3)=-\sqrt 3
\]
You should be able to finish it now

Answer 4

We have to find the solutions that are in the interval \( [0 ,2 \pi)\)

Answer 5

A moment of reflection reveals that
\[
\frac \pi 3\\
\frac \pi 3+\pi=\frac {4 \pi}3\\

-\frac \pi 3+\pi=\frac {2 \pi}3\\
-\frac \pi 3+2\pi=\frac {5 \pi}3\\

\]
I think that all the solutions.

Answer 6

Thanks you! Sorry I keep losing.connection and its not letting me reply

Answer 7

okay so tan^2x-3=0
You added 3 to both sides which turn out to be tan^2x= 3 then you square root both sides which make it cross the square root with tan^2
Which now equals tanx= square root of 3
Then you use the inverse trig function which would be x=inverse of tan square root of 3 + pi n
Which equals x= 1.0479 + 2 pi
And if you make a n and x table you get -2.0944 when n=1, 1.0479 when n= 0, 4.1888 when n=1 and 7.3304 when n=2 but since the interval is [0,2pi) the solutions are x=1.0479 and 4.1888 since 2pi= 6.2832 right??