1. ## solve for x

can someone show me how to solve this?

2. use the identity $\sec^2{x} = 1 + \tan^2{x}$ and get $\sec^2{x}$ in terms of tangent ... then solve the resulting factorable quadratic.

3. ## thanks!

thanks!

i don't know what to do after (tanx-3)(tanx+1)=0

help!

4. $ab = 0$. Either a = 0 or b = 0 or both.

So, set both factors equal to 0:
$\tan x - 3 = 0 \qquad \tan x - 1 = 0$

etc. etc. Just as you would if you had something like $(y-3)(y-1) = 0$.

5. ## okay

i got 71.57 and 45

are those right?

6. Yes .. but there are two more (in $[0, 2\pi]$ anyway).

ex. $\tan x = 1$

$x = 45^{\circ}$ works ... but look in quadrant III as well

Also, since adding multiples of 2pi (360 degrees) will give you the same spot on the unit circle, then you could generalize your answer to $x = 45^{\circ} + n360^{\circ}$ where n is an integer. This applies to your 4 other answers.