Question

Please solve:
5sin2x + 3sinx - 1 = 0

Answer 1

5(2sinxcosx) + 3sinx - 1 = 0
sinx(10cosx + 3) = 1
we know sinx = cos(90°-x) , so

cos(90-x)(10cosx + 3) = 1

10sinxcosx = 1 - 3sinx
100sin^2 x cos^2 x = 1 - 6sinx + 9sin^2 x
100sin^2 x (1 - sin^2 x) = 1 - 6sinx + 9sin^2 x
100sin^2 x - 100sin^4 x + 6sinx - 9sin^2x - 1 = 0

let sinx = y
91y^2 - 100y^4 + 6y - 1 = 0

nasty equation .....
I ran it through Wolfram and got
y = .0771 to be a verifies positive answer.
then sinx = .0771
x = 4.4219°

http://www.wolframalpha.com/input/?i=91y%5E2+-+100y%5E4+%2B+6y+-+1+%3D+0+

check:
LS = 5sin(8.84379°) + 3sin(4.4219) - 1
= 1.000005
close enough


so one solution is appr. 4.4219°

I had my calculator set in degrees, could just as well have it set in radians.

I did not check some of the other answers that Wolfram found.
Since I squared my equation at the start, all solutions would have to be verified.