Question

How do I solve the following equation step by step?

`-sin^2x=2cosx-2`

Answer 1

`x= 0+2pin` where `n∈Z`

Explanation:

`-sin^2x=2cosx-2`

1. Set the equation equal to 0
   `-sin^2x-2cosx+2=0`

2. Apply the modified Pythagorean identity: `1-cos^2theta=sin^2theta`:
   `-(1-cos^2theta)-2cosx+2=0`

3. Simplify by distributing
   `-1+cos^2x-2cosx+2=0`

4. Combine like terms
   `cos^2x-2cosx+1=0`

5. Factor like a quadratic
   `(cosx-1)(cosx-1)=0`

6. Solve
   `cosx=1`
   `x= 0+2pin` where `n∈Z`

graph{(sinx)^2+2cosx-2 [-0.92, 19.08, -5.48, 4.52]}