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2sin^2 theta + 3cos theta -3 =0 tyler, answer meee!! LOL
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\[\sin^2 \theta = 1 - \cos^2 \theta\]
\[2(1-\cos^2 \theta)- 3 \cos \theta -3 = 0\]
\[2-2\cos^2 \theta - 3 \cos \theta -3 = 0\]
\[-2 \cos^2 \theta - 3\cos \theta-1 = 0\]
times everything by -1 \[2\cos^2\theta + 3\cos \theta +1 = 0\]
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factor \[(2 \cos \theta + 1 )( \cos \theta + 1 ) = 0\]
solve the two parts... (2cosθ+1) = 0 or (cosθ+1)=0 cos θ =- .5 or cos θ = -1
θ = 120, 240, 180
-fin-
hmm...
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is that the pythagorean id?
\[\sin^2\theta=1−\cos^2\theta\] is Pythagorean's Identity
you meant -1 not equals 1.. right?
if you flip it around, \[1 = \sin^2\theta + \cos ^2\theta\]
ooh.
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