Question 772232: Find all the solutions to the equations in the interval (0, 2pi.
1. 2sin^2x=2+cosx
2. 2sin^2x+3sinx+1=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find all the solutions to the equations in the interval (0, 2pi.
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1. 2sin^2x=2+cosx
2(1-cos^2x)=2+cosx
2-2cos^2x=2+cosx
2cos^2x+cosx=0
cosx(2cosx+1)=0
..
cosx=0
no solution: (0 not in given domain)
..
2cosx+1=0
cosx=-1/2
x=2pi/3, 5pi/3
..
2. 2sin^2x+3sinx+1=0
solve for sinx using quadratic formula
a=2, b=3, c=1
ans:
sinx=-1
x=3pi/2
or
sinx=-1/2
x=7pi/6, 11pi/6
solutions:x=2pi/3, 5pi/3, 3pi/2, 7pi/6, 11pi/6
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