cos A + cos 2A + cos 3A = cos 2A(1-cos A)
nah, it is stated there to prove the identities/equations.
I have proved it actually but then I've realized that I was wrong and then I couldn't get the process anymore.
cos A + cos 2A + cos 3A = cos 2A(1-cos A)
What I did was to factor cos A, thus it makes cos A (1 + cos A + cos 2A)
then I've realized that it is wrong. Though I aririved at cos 2A(1-cos A)
reiward, Mr Fantastic is correct. What you have is a trigonometric equation, not an identity. His counter-example was when A = 0, that will give you 3 = 0, which is wrong. So it merits to be an equation, not an identity as what you claimed it to be. Se my graph . . . .
Are you meant to solve this for A (which is OK) or prove it, that is, show it's true for all A (which is not OK because it's not an identity). Chipce of words is important. Prove (the word you used) does not mean the same thing as solve (the word you might have meant ....)