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Hey can anyone help me? I'm trying to compute x in this equation:
18sinxcosx=1-6sinx
Would there be an easy way other than trial and error?
Thanks
18sinxcosx=1-6sinx
Would there be an easy way other than trial and error?
Thanks
Hmm..I posted something similar to that but deleted it since I set the coefficient of -a*sin(x) to 1 instead of 6.If you square both sides and plug in [itex]\cos^2 x = 1 - \sin^2 x[/itex] then you will obtain a quadratic in sin, which you can solve using the quadratic formula.
You dropped a '2' in that last equation: that should beBy trigonometric sine rule, sin(x)cos(x)=(1/2)[sin(x+x)+sin(a-x)]
Simplify this equation, we get (1/2)[sin(2x)+sin(x-x)]=sin(2x)/2 (*)
Plug (*) into the LHS, then we get 9sin(x)=1-6sin(x)
Solve, for x. It should be easy from here. =]
Alas, I looked at that possibility and it doesn't work for this equation. An approach of that sort will work for certain special cases of the numerical coefficients -- but this ain't one o'them...If you square both sides and plug in [itex]\cos^2 x = 1 - \sin^2 x[/itex] then you will obtain a quadratic in sin, which you can solve using the quadratic formula.
Ah, sorry. I didn't see that.You dropped a '2' in that last equation: that should be
9 sin(2x) = 1 - 6 sin(x) ,
which is what I gave back in post #2. Unfortunately, there isn't any nice way to solve that...