SOLUTION: sin^2x-cos^2x=cosx between intervals (o,2pi)

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Question 1027900: sin^2x-cos^2x=cosx
between intervals (o,2pi)
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
%281-cos%5E2%28x%29%29-cos%5E2%28x%29=cos%28x%29

1-2cos%5E2%28x%29-cos%28x%29=0

2cos%5E2%28x%29%2Bcos%28x%29-1=0

cos%28x%29=%28-1%2B-+sqrt%281-4%2A2%2A%28-1%29%29%29%2F%282%2A2%29

cos%28x%29=%28-1%2B-+sqrt%289%29%29%2F4

cos%28x%29=%28-1%2B-+3%29%2F4

system%28cos%28x%29=-1%2COR%2Ccos%28x%29=1%2F2%29

system%28x=180%2COR%2Cx=60%2COR%2Cx=300%29------degrees