SOLUTION: Find all the values of x in [0,2π] that satisfy (cosx)^2 −3sinx−3=0.

Algebra ->  Trigonometry-basics -> SOLUTION: Find all the values of x in [0,2π] that satisfy (cosx)^2 −3sinx−3=0.      Log On


   

Question 983895: Find all the values of x in [0,2π] that satisfy (cosx)^2 −3sinx−3=0.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Use the identity cos^2 = 1 - sin^2 to go from this

cos%5E2%28x%29+-+3sin%28x%29+-+3+=+0

to this

1+-+sin%5E2%28x%29+-+3sin%28x%29+-+3+=+0

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Now combine like terms to get -sin%5E2%28x%29+-+3sin%28x%29+-+2+=+0

Let z+=+sin%28x%29 which gives this new equation -z%5E2+-+3z+-+2+=+0

Solve the equation -z%5E2+-+3z+-+2+=+0 for z, using the quadratic formula. There are 2 solutions for z and they are z+=+-1 or z+=+-2

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Recall that z+=+sin%28x%29 so saying z+=+-1 or z+=+-2 really means

sin%28x%29+=+-1 or sin%28x%29+=+-2

I'll let you solve from here. Use the unit circle. Hint: one of those equations has no solution.