SOLUTION: Find the solutions of the equation that are in the interval [0,2π). Please show steps (especially rearranging the equation to solve it. Thanks for helping me. I know the solu
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Question 148653: Find the solutions of the equation that are in the interval [0,2π). Please show steps (especially rearranging the equation to solve it. Thanks for helping me. I know the solution is 7pi/6, 11pi/6, pi/2.
sinb+2cos^(2)b=1
Answer by Nate(3500) (Show Source): You can put this solution on YOUR website!
sin(b) + 2cos^2(b) = 1
sin(b) + 2(1 - sin^2(b)) = 1
sin(b) + 2 - 2sin^2(b) = 1
2sin^2(b) - sin(b) - 1 = 0
* when you factor, you treat this much like any other polynomial .. for understanding sakes, x = sin(b)
2x^2 - x - 1 = 0
2x^2 - 2x + x - 1 = 0
(2x^2 - 2x) + (x - 1) = 0
2x(x - 1) + (x - 1) = 0
(2x + 1)(x - 1) = 0
* x = sin(b)
(2sin(b) + 1)(sin(b) - 1) = 0
2sin(b) + 1 = 0 and sin(b) - 1 = 0
sin(b) = -1/2 and sin(b) = 1
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