Question 533337: Find all the solutions in the interval (0,2pi).
square root of 2 times sinx secx = 2sinx
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find all the solutions in the interval (0,2pi).
square root of 2 times sinx secx = 2sinx
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√2sinx secx=2sinx
√2sinx secx-2sinx=0
sinx(√2secx-2)=0
..
sinx=0
x=arcsin(0)=0 and π
..
√2secx-2=0
secx=2/√2=1/cosx
cosx=√2/2
x=arccos(√2/2)=π/4 and 7π/4 (in quadrants I and IV where cos and sec>0)
solutions: 0, π/4, π, and 7π/4
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