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.
Can you help me solve:
5sin2x + 3sinx - 1 = 0
Thanks
2 answers
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
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
1 answer