Question

2cos²x+2cosx+sin²x=0

Answer 1

2cos²x+2cosx+sin²x=0

2cos²x+2cosx+1-cos²x=0

cos²x+2cosx+1=0

( cosx + 1 )²=0

cosx=-1

x=π+2πn, n € Z

Answer 2

To solve the equation 2cos²x + 2cosx + sin²x = 0, we can use the trigonometric identity sin²x + cos²x = 1. Rewrite the equation in terms of sine and cosine:

2(1 - sin²x) + 2cosx + sin²x = 0

Now, distribute the 2 on the left side of the equation:

2 - 2sin²x + 2cosx + sin²x = 0

Combine like terms:

2cosx - sin²x = -2

Now, you can rearrange the equation:

2cosx = -2 + sin²x

Divide both sides by 2:

cosx = (-2 + sin²x)/2

cosx = -1 + (1 - sin²x)/2

cosx = -1 + (1/2) - (sin²x/2)

cosx = -1/2 - (sin²x/2)

Now, you have cosx in terms of sinx:

cosx = -1/2 - (sin²x/2)

You can use this equation to find the values of x for which it holds true. Depending on the context or the range of x you're interested in, you may need to further simplify or manipulate the equation to find specific solutions.