Question
Given that 3cos2x + 17sinx = 8, find the exact value of sinx .
Answers
oobleck
3cos2x + 17sinx - 8 = 0
3(1-2sin^2x) + 17sinx - 8 = 0
6sin^2x - 17sinx + 5 = 0
(3sinx-1)(2sinx-5) = 0
sinx = 1/3 or 2/5
3(1-2sin^2x) + 17sinx - 8 = 0
6sin^2x - 17sinx + 5 = 0
(3sinx-1)(2sinx-5) = 0
sinx = 1/3 or 2/5
R_scott
cos(2x) = 1 – 2 sin^2(x)
substituting ... 3 - 6 sin^2(x) + 17 sin(x) = 8
... -6 sin^2(x) + 17 sin(x) - 5 = 0
... 6 sin^2(x) - 17 sin(x) + 5 = 0
factoring ... [3 sin(x) - 1] [2 sin(x) - 5] = 0
3 sin(x) - 1 = 0 ... sin(x) = 1/3
2 sin(x) - 5 = 0 ... sin(x) = 5/2 ... unrealistic solution
substituting ... 3 - 6 sin^2(x) + 17 sin(x) = 8
... -6 sin^2(x) + 17 sin(x) - 5 = 0
... 6 sin^2(x) - 17 sin(x) + 5 = 0
factoring ... [3 sin(x) - 1] [2 sin(x) - 5] = 0
3 sin(x) - 1 = 0 ... sin(x) = 1/3
2 sin(x) - 5 = 0 ... sin(x) = 5/2 ... unrealistic solution