solve the given equation over the interval [0,2pi): 2cos^2(x)-2cos(x)+squareroot 2+cos(x)-squareroot 2=0

Answer 1

2cos^2(x) - 2cos(x) + sqr2+cosx -sqr2 = 0

2cos^2(x) -cosx = 0

cosx(2cosx -1) = 0

cosx=0 or 1/2

x = 0, pi/3 or 5pi/3 radians

or x= 0, 60 or 300 degrees

which seems to be a relatively "simple" solution

UNLESS you meant to write

2cos^2(x) -2cosx +sqr(2+cosx)-sqr2 = 0

then you get a different answer

it seems to devolve into an 8th degree equation with 6, 4, 2 or 0 possible real solutions, max 5 positive real, max 1 negative real solution. multiplying twice by a variable may introduce extraneous solutions

16cos^8(x)-64cos^7(x)+6cos^6(x)-4cos^5(x)+16cos^4(x)-8cosx -16 =0

which isn't easy to solve, but by DesCartes' Method, anywhere from 0 to 6 possible solutions

even if there are solutions, you'd need to check them with the original equation to see if they work. Given the complexity, best guess is go with the original solution: 0, pi/3, and 5pi/3. UNLESS there's a different misprint or mistake in the original problem.