Help to solve cos^2 x-1=0 for 0≤ x< 2pi?

cos x = ± 1
x = 0 , π

First let's add 1 to both sides of the equation to isolate cos^2 x


cos^2 x - 1 = 0

cos^2 x - 1 + 1 = 0 + 1

cos^2 x = 1 <-- now we square root both sides of the equation...this is the inverse of squaring..

√(cos^2 x) = √(1) <-- remember this can be plus or minus...

cos x = ±1 <-- now look at the unit circle for this...the x-axis...


Where cos x = 1 is at 0 or 2π (but we do not consider 2π as this solution is NOT in the domain given as it says LESS than 2π but not LESS THAN OR EQUAL to 2π), so just 0.

Where cos x = -1 is at π...so our solutions are:



x = 0, π <-- answer...