Question

2cosA+cos3A+cos5A=4cosA*cos^{2} 2A

Answer 1

Cos 3A= 4Cos^3A-3CosA. This is an identity.
Cos(A+2A)= CosACos2A-SinASin2A. -(1)
Cos2A=Cos^2A-Sin^2A and Sin2A= 2SinACosA
Put this in equation (1) and you'll get the identity.
So the equation
Cos3A+CosA-2Cos2A =  4 Cos^3A- 2CosA- 2(Cos^2A-Sin^2A)
                                      =4 Cos^3A-2CosA-2(2Cos^2A-1)
                                       =4Cos^3A-2CosA-4Cos^2A+2
So, 2CosA-4Cos^2A-4Cos^3A=2
Now assume CosA=x
So 2x+4x^2-4X^3=2 or 4X^3-4X^2-2x+2=0
Find X. 
And CosA=X
So A= Cos ^-1(X)
Edit: For further solving the quadratic equation 
4X^3-4x^2-2x+2=0
2x^3-2x^2-x+1=0
X^3-x^2-x/2+1/2=0
Club x^3 and x^2 together 
So x^2(x-1)-1/2(x-1)=0
So we've got two factors 
(X-1)(x^2-1/2)=0