Posted by Sally on Tuesday, December 20, 2011 at 11:57pm.
2 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)
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(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.
In #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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