SOLUTION: Find the solutions for 2sin^2(2x)+5sin(2x)-3=0

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Question 377998: Find the solutions for 2sin^2(2x)+5sin(2x)-3=0
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = 2 sin (2x). This expression becomes 2a%5E2+%2B+5a+-+3+=+0. The equation factors to (2a - 1)(a + 3) = 0, so we have a = 1/2 or a = 3. Therefore we have
2 sin (2x) = 1/2 --> sin (2x) = 1/4
2 sin (2x) = -3 --> sin (2x) = -3/2
The second equation has no solutions for x, as the sine function is defined with outputs [-1,1]. The first one has solution 2x+=+arcsin%281%2F4%29 or
x+=+%28arcsin%281%2F4%29%29%2F2
Note that we may add any integer multiple of pi to x, as sine is periodic with period 2pi (note that x is multiplied by 2).