Factorising Equation for dI/dθ = 0 with I = 0.8x10^-5 and Cosine Functions

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I want [tex]\frac{dI}{d \theta} = 0[/tex] when [tex]I = 0.8 \cdot 10^{-5}[\frac{273}{16} + \frac{17}{2}cos(kd sin \theta) + 2cos(2kd sin \theta)][/tex]

If I've calculated correctly, this means that [tex]-8.5sin(8.1sin \theta)8.1cos \theta - 32.4sin(16.2sin \theta)cos \theta = 0[/tex]. Can somebody help me factorise this?
 
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If only the stuff in brackets is the argument of the first sine of each term, you could get away with dividing through by [tex]cos \theta[/tex] to leave you with

[tex] -8.5sin(8.1sin \theta)8.1 - 32.4sin(16.2sin \theta) = 0[/tex]

which is equivalent to
[tex] -68.85sin(8.1sin \theta) - 32.4sin(16.2sin \theta) = 0[/tex]
 
Then recognising 16.2sinθ = 2(8.1sinθ) we can apply the sin double angle formula (sin2α = 2cosαsinα) on the second term