Please solve:
5sin2x + 3sinx - 1 = 0
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.
To solve the equation 5sin2x + 3sinx - 1 = 0, we can use a substitution to simplify it. We'll use a substitution such that sinx = t, and rewrite the equation in terms of t.
Let's replace sinx with t:
5sin2x + 3sinx - 1 = 0
5(2sinxcosx) + 3sinx - 1 = 0 (using the double-angle identity: sin2x = 2sinxcosx)
10sintcost + 3sinx - 1 = 0
Now, we have an equation in terms of a single variable t:
10t(1 - t^2) + 3t - 1 = 0
Expanding and rearranging the equation:
10t - 10t^3 + 3t - 1 = 0
-10t^3 + 13t - 1 = 0
We have a cubic equation in terms of t. To solve this equation, we can use numerical methods or graphical methods. One commonly used method is the Newton-Raphson method or using a graphing calculator.
However, if you are looking for an exact solution, finding the roots of a cubic equation can be quite complex. In this case, you may need to use numerical methods to approximate the values of t.
Once you have the values of t, you can substitute them back into sinx = t to find the corresponding values of x.