Question 674356
Given: cosθ= -3/5 and sinθ > 0. 
You are working with a 3-4-5 right triangle and reference angle in quadrant II where cos<0 and sin>0
cosx=-3/5=adj side/hypotenuse
opp side=&#8730;(5^2-3^2)=&#8730;(25-9)=&#8730;16=4
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Find the exact value of cos x/2.
Identity: cos (x/2)=±&#8730;[(1+cosx)/2]
=-&#8730;[(1-3/5)/2]
=-&#8730;[(2/5)/2]
=&#8730;(2/10)
=&#8730;(1/5)
=1/&#8730;5
=&#8730;5/5
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solve cos 2x=(sqrt2)/2 on the interval [0,2pi). Four solutions.
cos 2x=&#8730;2/2
2x=&#960;/4, and 7&#960;/4 (in quadrants I and IV where cos>0) (2 solutions)
x=&#960;/8 and 7&#960;/8
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solve cos^(2)x+2cos+1=0 on the interval [0, 2pi) 
cos^(2)x+2cos+1=0
(cosx+1)^2=0
cosx=-1
x=&#960; (multiplicity 2)