Solving trig functions! urgent please help!
posted by Sally .
1.2sinxtanx + tanx2sinx1=0
2.6tan^2x4sin^2x=1
Both functions is for 0<x<2pi
thank you!

2 sin^2x/cosx + sinx/cosx  2 sinx  1 = 0
2 sin^2x + sinx 2 sinxcosx  cosx = 0
2sinx(sinxcosx) = (sinxcosx)
sinx = 1/2
x = 210 degrees(7pi/6) or 330degrees(11pi/6) 
6 (sin^2/cos^2) 4 sin^2  1 = 0
6 s^2  4 s^2c^2  c^2 = 0
6 s^2 4s^2(1s^2) (1s^2) = 0
6s^2 4s^2 +4s^4 +s^2  1 = 0
4 s^4 +3 s^2 1 = 0
(4s^21)(s^2+1) = 0
sin^2x = 1/4
or
sin^2x = 1 (imaginary root)
sin x = +1/2
or sin x = 1/2
You can take it from there. 
In #1, Damon lost 2 answers by just dropping the (sinxcosx)
from
2sinx(sinxcosx) = (sinxcosx)
2sinx(sinxcosx) + (sinxcosx)= 0
(sinxcosx)(2sinx + 1) = 0
sinx = 1/2
etc, (see above)
or
sinx  cosx = 0
sinx = cosx
sinx/cosx =1
tanx = 1
x = 45° or 225° (π/4 or 5π/4)
so x = π/4, 5π/4 , 7π/6, 11π/6