2sinxcosx+cosx=0

To solve the equation 2sin(x)cos(x) + cos(x) = 0, we can factor out cos(x) from both terms:

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

Now we have two possibilities:

1. cos(x) = 0:
To solve this, we need to find the values of x where cos(x) = 0. This occurs at x = π/2 + nπ, where n is an integer.

2sin(x) + 1 = 0:
Subtracting 1 from both sides:
2sin(x) = -1

Dividing both sides by 2:
sin(x) = -1/2

To find the values of x where sin(x) = -1/2, we can look at the unit circle or use the reference angle of 30 degrees. The solutions are x = 7π/6 + 2nπ and x = 11π/6 + 2nπ, where n is an integer.

So the solutions to the original equation 2sin(x)cos(x) + cos(x) = 0 are x = π/2 + nπ (for cos(x) = 0) and x = 7π/6 + 2nπ or x = 11π/6 + 2nπ (for sin(x) = -1/2).