Factor 2sinxcosx+cosx=0

To factor the expression 2sin(x)cos(x) + cos(x) = 0, we can first notice that there is a common factor of cos(x) in both terms. We can factor out cos(x) to get:

cos(x)(2sin(x) + 1) = 0

Now, we need to solve for the values of x that make the expression equal to zero. We have two possibilities:

1. cos(x) = 0
2. 2sin(x) + 1 = 0

1. When cos(x) = 0, x = π/2 + nπ where n is an integer.

2. When 2sin(x) + 1 = 0, we can solve for sin(x):

2sin(x) + 1 = 0
2sin(x) = -1
sin(x) = -1/2

This occurs at x = 7π/6 + 2nπ and x = 11π/6 + 2nπ, where n is an integer.

Therefore, the solutions to the equation 2sin(x)cos(x) + cos(x) = 0 are x = π/2 + nπ, 7π/6 + 2nπ, and 11π/6 + 2nπ where n is an integer.