Question

I need precal help please!!

Answer 1

post your question please

Answer 2

one second

Answer 3

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

Answer 4

so then how do i find the solutions

Answer 5

you will get tan2x= \[2\sqrt{2}\]

Answer 6

so the points on the unit circle would be what

Answer 7

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

Answer 8

none of those are the selected answers?

Answer 9

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

Answer 10

i dont know how to do that?

Answer 11

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

Answer 12

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

Answer 13

it's d

Answer 14

how did they get pi?

Answer 15

that was wrong.....thanks

Answer 16

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)