The equation
$$3\sin^2 x - 3\cos x -6\sin x + 2\sin 2x + 3=0$$
has a solution $x = 0$. That means it has a factor of $\cos x - 1$. I tried to write the given equation has the form
$$(\cos x - 1)P(x)=0.$$
I am looking for the factor $P(x)$. How can I get Mathematica to find it?
TrigFactor[3 Sin[x]^2 - 3 Cos[x] - 6 Sin[x] + 2 Sin[2 x] + 3]
gives the output-2 Sqrt[2] Sin[\[Pi]/4 - x/2] (Cos[x/2] - 3 Sin[x/2]) (2 Cos[x/2] - Sin[x/2]) Sin[x/2]
. This has an overall factorSin[x/2]
which gives yourx = 0
solution. $\endgroup$ – Stephen Luttrell Mar 31 '13 at 15:11