SOLUTION: Solve each equation on the interval (0,2pi): a.) 2sin^(2)x-cos x-1=0, b.) sin 3x cos 2x + cos3x sin2x= (√2/2).
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Question 685127: Solve each equation on the interval (0,2pi): a.) 2sin^(2)x-cos x-1=0, b.) sin 3x cos 2x + cos3x sin2x= (√2/2).
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Solve each equation on the interval (0,2pi): a.) 2sin^(2)x-cos x-1=0, b.) sin 3x cos 2x + cos3x sin2x= (√2/2).
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a.) 2sin^(2)x-cosx-1=0
2(1-cos^2x)-cosx-1=0
2-2cos^2x-cosx-1=0
2cos^2x+cosx-1=0
(2cosx-1)(cosx+1)=0
2cosx-1=0
cosx=1/2
x=π/3, 5π/3
or
cosx+1=0
cosx=-1
x=π
..
b.) sin 3x cos 2x + cos3x sin2x= (√2/2).
Identity:sin(s+t)=sin s cos t+cos s sin t
sin 3x cos 2x + cos3x sin2x= (√2/2)
sin(3x+2x)=sin(5x)=√2/2
5x=π/4, 3π/4
x=π/20, 3π/20
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