Question 674356: Given: cosθ= -3/5 and sinθ > 0. Find the exact value of cos θ/2.
solve cos 2x=(sqrt2)/2 on the interval [0,2pi). Four solutions.
solve cos^(2)x+2cos+1=0 on the interval [0, 2pi)
if you could show me how you solved it, it would help a lot. thank you.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 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=√(5^2-3^2)=√(25-9)=√16=4
...
Find the exact value of cos x/2.
Identity: cos (x/2)=±√[(1+cosx)/2]
=-√[(1-3/5)/2]
=-√[(2/5)/2]
=√(2/10)
=√(1/5)
=1/√5
=√5/5
..
solve cos 2x=(sqrt2)/2 on the interval [0,2pi). Four solutions.
cos 2x=√2/2
2x=π/4, and 7π/4 (in quadrants I and IV where cos>0) (2 solutions)
x=π/8 and 7π/8
..
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=π (multiplicity 2)
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