solve 2cos²x+3cosx=-2 for all values in between 0 and 360 degrees

Question

can't factor this

Answer 1

If we use the quadratic formula for solving cos(x) we get an imaginary root: √(9-16)=√-7.

However, if -2 is replaced by 2:

2cos²(x)+3cos(x)-2=0=(2cos(x)-1)(cos(x)+2), from which cos(x)=½. cos(x) cannot exceed 1 or be less than -1, so we can reject cos(x)=-2.

Therefore x=π/3 (60°) or -π/3, which we can shift by adding 2π, x=5π/3 (300°).