Given: cos a 2cos b cos c = 2
To find: The relation between a, b, and c

Solution:
We know that cos x = cos(-x)
So, we can write the given expression as:

cos a - 2cos b cos c = -2

Adding 2cos b cos c on both sides, we get:

cos a = 2cos b cos c + 2

Using the identity cos (x+y) = cos x cos y - sin x sin y, we can write:

cos (b+c) = cos b cos c - sin b sin c

Multiplying both sides by -2, we get:

-2cos b cos c + 2sin b sin c = -2cos (b+c)

Adding 2 on both sides, we get:

-2cos b cos c + 2sin b sin c + 2 = -2cos (b+c) + 2

Using the identity cos (x+y) = cos x cos y - sin x sin y, we can write:

-2cos b cos c + 2sin b sin c + 2 = -2cos b cos c + 2sin b cos c + 2cos b

Simplifying, we get:

2sin b cos c + 2cos b = cos a

Substituting the value of cos a from the given expression, we get:

2sin b cos c + 2cos b = 2cos b cos c + 2

Simplifying, we get:

2(cos b - 1)(cos c - sin b) = 0

So, either cos b = 1 or cos c = sin b

Case 1: cos b = 1
This implies b = 2πn, where n is an integer.
Substituting this in the expression for cos a, we get:

cos a = 2cos c + 2

Using the identity cos^2 x + sin^2 x = 1, we can write:

cos^2 c + sin^2 c = 1

Substituting cos^2 c = 1 - sin^2 c, we get:

2cos^2 c - 2 = cos a

Dividing by 2, we get:

cos^2 c - 1 = -cos(a/2)^2

Using the identity sin^2 x + cos^2 x = 1, we can write:

sin^2 (a/2) = cos^2 c

Substituting in the previous expression, we get:

sin^2 (a/2) - 1 = -cos(a/2)^2

Using the identity sin^2 x - cos^2 x = -cos 2x, we can write:

cos (a/2) = ±√(1-cos a)/2

Substituting the value of cos a, we get:

cos (a/2) = ±√(1-2cos b cos c)/2

Case 2: cos c = sin b
Using the identity sin^2 x + cos^2 x = 1, we can write:

sin^2 b + cos^2 c = 1

Substituting cos c = sin b, we get:

2cos^2 b - 1 = 0

This

A
2b=a+c