Solve the equation

cos(2x)cos(x)+sin(2x)+sin(x)=1 for all values of x.

To solve the equation cos(2x)cos(x) + sin(2x) + sin(x) = 1, we can use some trigonometric identities to simplify it. Let's break it down step by step:

First, we can use the double angle identity for cosine: cos(2x) = 2cos^2(x) - 1. Substituting this in, our equation becomes:

(2cos^2(x) - 1)cos(x) + sin(2x) + sin(x) = 1

Next, let's use the double angle identity for sine: sin(2x) = 2sin(x)cos(x). Substituting this in, our equation becomes:

(2cos^2(x) - 1)cos(x) + 2sin(x)cos(x) + sin(x) = 1

Now, let's simplify further by rearranging the terms:

2cos^3(x) + sin(x) + 2sin(x)cos(x) - cos(x) = 1

Grouping the terms with cos(x) and sin(x):

2cos^3(x) + (2sin(x)cos(x) - cos(x)) + sin(x) = 1

Factoring out cos(x) from the second term:

2cos^3(x) + cos(x)(2sin(x) - 1) + sin(x) = 1

Now, let's focus on the second term inside the parentheses, i.e., 2sin(x) - 1:

For this term to be equal to zero, we have: 2sin(x) - 1 = 0
Solving this equation gives us sin(x) = 1/2.
This means that x can take on two possible values: x = π/6 + 2πn or x = 5π/6 + 2πn, where n is an integer.

Checking these values in the original equation, we find that for x = π/6 + 2πn, the equation holds true. Similarly, for x = 5π/6 + 2πn, the equation holds true as well.

Therefore, the solution to the equation cos(2x)cos(x) + sin(2x) + sin(x) = 1 for all values of x is:
x = π/6 + 2πn, where n is an integer,
and x = 5π/6 + 2πn, where n is an integer.