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The reason for Wolfram's astonishingly low term count for the OP example, is that the example given is a special case. Because the initial sin/cos arguments are small integers, the set of unique argument sums as computed in the product-to-sum conversions is a very small set--- just 59 unique arguments, to be exact. So for the OP example, the answer is ...


This is not a reply where your error lies, but we can solve the equation for $x$ by hand because the equation can be written as $$0 = \frac{acdfF(x)}{A}+bcfhj\left(1-\frac{F(x)}{A}\right)$$ where $$F(x)=bj(bj-cfhx)-ad(bj+cfix)$$ $$A=a^2d^2-abd(g+1)j+b^2j^2.$$ Then, multiplying the both sides by $A$ gives $$0=acdfF(x)+bcfhj(A-F(x)).$$ Dividing the both sides ...


Don't know about Wolfram, but I put your formula in Mathematica as Solve[0 == a*c*d*f*(b*j*(b*j - c*f*h*x) - a*d*(b*j + c*f*i*x))/(a^2*d^2 - a*b*d*(g + 1)*j + b^2*j^2) + b*c*f*h*j*(1 - (b*j*(b*j - c*f*h*x) - a*d*(b*j + c*f*i*x))/(a^2*d^2 - a*b*d*(g + 1)*j + b^2*j^2)), x] // Simplify And got the answer x -> (a b d j (a d (-1 + h) + b (1 ...


Note how you have defined $\sinh(t)$. A factor of $\frac{1}{2}$ is missing, which when you correct for should give you the answer you're looking for.

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