SOLUTION: 2sin^2x=3sinx+5 solve equation on interval [0,2pi)

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Question 1001690: 2sin^2x=3sinx+5
solve equation on interval [0,2pi)
Answer by ikleyn(52847) About Me  (Show Source):
You can put this solution on YOUR website!
.
2%2Asin%5E2%28x%29 = 3%2Asin%28x%29 + 5.

Let us introduce a new variable,  y = sin(x).
Then the equation takes the form

2y%5E2 = 3y+%2B+5,     or

2y%5E2 - 3y - 5 = 0.

Solve it by using the quadratic formula:

y%5B1%2C2%5D = %283+%2B-+sqrt%283%5E2+%2B+4%2A2%2A5%29%29%2F4 = %283+%2B-+sqrt%2849%29%29%2F4 = %283+%2B-+7%29%2F4.

y%5B1%5D = -1   ----->   sin(x) = -1   ----->   x = 3%2Api%2F2.

y%5B2%5D = 10%2F4  doesn't fit,  because  sin(x)  can not be more than  1.

Answer.  x = 3%2Api%2F2.