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September 30, 2014

September 30, 2014

Posted by **Sally** on Tuesday, December 20, 2011 at 11:57pm.

2.6tan^2x-4sin^2x=1

Both functions is for 0<x<2pi

thank you!

- Solving trig functions! urgent please help! -
**Damon**, Wednesday, December 21, 2011 at 5:29am2 sin^2x/cosx + sinx/cosx - 2 sinx - 1 = 0

2 sin^2x + sinx -2 sinxcosx - cosx = 0

2sinx(sinx-cosx) = -(sinx-cosx)

sinx = -1/2

x = 210 degrees(7pi/6) or 330degrees(11pi/6)

- Solving trig functions! urgent please help! -
**Damon**, Wednesday, December 21, 2011 at 5:46am6 (sin^2/cos^2) -4 sin^2 - 1 = 0

6 s^2 - 4 s^2c^2 - c^2 = 0

6 s^2 -4s^2(1-s^2) -(1-s^2) = 0

6s^2 -4s^2 +4s^4 +s^2 - 1 = 0

4 s^4 +3 s^2 -1 = 0

(4s^2-1)(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.

- Solving trig functions! urgent please help! -
**Reiny**, Wednesday, December 21, 2011 at 8:57amIn #1, Damon lost 2 answers by just dropping the (sinx-cosx)

from

2sinx(sinx-cosx) = -(sinx-cosx)

2sinx(sinx-cosx) + (sinx-cosx)= 0

(sinx-cosx)(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

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