Can you help me solve:
5sin2x + 3sinx - 1 = 0
Thanks
this doesn't look easy. I'd attack it like this:
10sinx cosx + 3sinx - 1 = 0
sinx(10cosx+3) = 1
sinx = 1/(10cosx+3)
sin^2x = 1/(100cos^2x + 60cosx + 9)
(1-cos^2x)(100cos^2x+60cosx+9) - 1 = 0
Expand that out, and you have a 4th-degree polynomial in cosx. Guaranteed not to be easy.
This one is probably best tackled with a numeric method, starting with the graph for initial estimates:
http://www.wolframalpha.com/input/?i=5sin2x+%2B+3sinx+-+1
Now, if there was a typo, and you meant
5sin^2(x) + 3sin(x) - 1 = 0
then things are easier, but the answer is not any well-known angles.
I had actually followed the same method Steve suggested.
My first 4 lines were an alternate attempt, which does not fit in with the rest of the solution. I should have deleted those 4 lines.
As Steve pointed out it was a mess, and I had to use Wolfram
http://www.jiskha.com/display.cgi?id=1452459531
Of course! To solve the equation 5sin2x + 3sinx - 1 = 0, we can use the substitution u = sinx.
Step 1: Substitute u = sinx into the equation.
5sin2x + 3sinx - 1 = 0
5(2u^2) + 3u - 1 = 0
10u^2 + 3u - 1 = 0
Step 2: Solve the quadratic equation.
To solve the quadratic equation 10u^2 + 3u - 1 = 0, we can use the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 10, b = 3, and c = -1.
u = (-3 ± √(3^2 - 4(10)(-1))) / (2(10))
u = (-3 ± √(9 + 40)) / 20
u = (-3 ± √49) / 20
u = (-3 ± 7) / 20
So we have two possible values for u:
u1 = (7 - 3) / 20
u1 = 4 / 20
u1 = 0.2
u2 = (-7 - 3) / 20
u2 = -10 / 20
u2 = -0.5
Step 3: Solve for x.
Now that we have the values of u, we can use the inverse sine function to solve for x.
For u = 0.2:
sinx = 0.2
x = arcsin(0.2) + 2πn or x = π - arcsin(0.2) + 2πn, where n is an integer.
For u = -0.5:
sinx = -0.5
x = arcsin(-0.5) + 2πn or x = π - arcsin(-0.5) + 2πn, where n is an integer.
By substituting these values back into the original equation, you can verify if they satisfy the equation.