I need precal help please!!

Question

ghazi · Answer

post your question please

anonymous · Answer

one second

anonymous · Answer

attachment

ghazi · Answer

tanx= sinx/cosx and 1-tan^2x = sec^2 x so $$2tanx/\sec ^{2}x= 2 tanx * \cos ^{2} x/\sin ^{2} x $$

anonymous · Answer

so then how do i find the solutions

ghazi · Answer

you will get tan2x= $$2\sqrt{2}$$

anonymous · Answer

so the points on the unit circle would be what

ghazi · Answer

the general solution of this will be (30+180n) or (pi/6+ pi*n)

anonymous · Answer

none of those are the selected answers?

ghazi · Answer

okay one more alternate solution to your problem....just take y= tan^2x and solve the quadratic you're gonna get general solution

anonymous · Answer

i dont know how to do that?

ghazi · Answer

just put tanx= y you will get 2*y= 1-y^2 rearranging the equation you'll get y^2+2y-1=0 now you can solve for y and just substitute the solution of y= tanx

anonymous · Answer

thats not one of the answers they gave!! look at the attachment

ghazi · Answer

it's d

anonymous · Answer

how did they get pi?

anonymous · Answer

that was wrong.....thanks

ghazi · Answer

none of the option seems correct for tanx=1 and pi is included for a general solution and for in third quadrant it's tan45=1 (tan is positive)