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

[kasse]

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

If I've calculated correctly, this means that -8.5sin(8.1sin \theta)8.1cos \theta - 32.4sin(16.2sin \theta)cos \theta = 0. Can somebody help me factorise this?

 

[joeyar]

If only the stuff in brackets is the argument of the first sine of each term, you could get away with dividing through by cos \theta to leave you with

<br /> -8.5sin(8.1sin \theta)8.1 - 32.4sin(16.2sin \theta) = 0<br />

which is equivalent to
<br /> -68.85sin(8.1sin \theta) - 32.4sin(16.2sin \theta) = 0<br />

 

[joeyar]

Then recognising 16.2sinθ = 2(8.1sinθ) we can apply the sin double angle formula (sin2α = 2cosαsinα) on the second term