SOLUTION: What are the x's of: ((sin^2)x) +(tan)x) -1)) = 0

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Question 747138: What are the x's of: ((sin^2)x) +(tan)x) -1)) = 0
Answer by KMST(5397) About Me  (Show Source):
You can put this solution on YOUR website!
%28sin%28x%29%29%5E2%2Btan%28x%29-1=0 <--> 1-%28cos%28x%29%29%5E2%2Btan%28x%29-1=0 <--> -%28cos%28x%29%29%5E2%2Btan%28x%29=0 <--> %28cos%28x%29%29%5E2=tan%28x%29 <--> %28cos%28x%29%29%5E2=sin%28x%29%2Fcos%28x%29 <--> %28cos%28x%29%29%5E3=sin%28x%29
%28cos%28x%29%29%5E3=sin%28x%29 --> %28cos%28x%29%29%5E6=%28sin%28x%29%29%5E2 --> %28cos%28x%29%29%5E6=1-%28cos%28x%29%29%5E2
Calling %28cos%28x%29%29%5E2=y we can re-write that equation as
y%5E3=1-y <--> y%5E3%2By-1=0
The only real solution is approximately y=0.6823278038
Going back to x we have %28cos%28x%29%29%5E2=0.6823278038
Between 0%5Eo and 360%5Eo (between 0 and 2pi radians) there are 4 angles with %28cos%28x%29%29%5E2=0.6823278038.
However, solutions to %28cos%28x%29%29%5E2=tan%28x%29 require positive tangent, so the only solutions will be in the first and third quadrants.
x=34.3%5Eo (or x=0.598767 radians) is the approximate solution in the first quadrant.
x=34.3%5Eo%2B180%5Eo=214.3%5Eo (or x=0.598767%2Bpi=3.740539 radians) is the approximate solution in the third quadrant.