SOLUTION: [(Sqrt2) secx-2](2sinx+1)=0 Find all solutions for this equation with the interval [0, 2pi)

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Question 707588: [(Sqrt2) secx-2](2sinx+1)=0
Find all solutions for this equation with the interval [0, 2pi)
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
[(Sqrt2) secx-2](2sinx+1)=0
Find all solutions for this equation with the interval [0, 2pi)
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√2secx-2=0
√2secx=2
secx=2/√2
cosx=√2/2
x=π/4, 7π/4 ( In quadrants I and IV where sec and cos>0)