Solving sin(x)/sin(x+215) = ab/cd

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To solve the equation sin(x)/sin(x+215) = ab/cd, one can utilize the identity that relates sin(x) and sin(x+a). The equation can be simplified to the form sin(x)(1-bcos(a)) = bcos(x)sin(a). This leads to the expression tan(x) = bsin(a)/(1-bcos(a)), allowing for the use of the inverse tangent function to find x. The discussion highlights the transformation of the original equation into a solvable format. The approach effectively demonstrates how to manipulate trigonometric identities for problem-solving.
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This isn't homework, but I can't seem to access the math forum at the moment, so I thought I'd ask here.

Homework Statement


sin(x)/sin(x+215) = ab/cd.
Solve for x

Homework Equations


Is there any property for sin x / sin y so I can simplify this?
 
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Your equation has the form sin(x) = bsin(x+a) = bsin(x)cos(a) + bcos(x)sin(a)
= bsin(x)(cos(a) + bsin(a))

sin(x)(1-bcos(a)) = bcos(x)sin(a)
\frac{\sin(x)}{\cos(x)}=\tan(x)=\frac{b\sin(a)}{1-b\cos(a)}

You can now solve using the inverse tangent function.
 
Thanks LCKurtz!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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