Question

How do you find all the solutions in the interval [0, 2pi): `2 cos^2(2x) - 1 = 0`?

Answer 1

Isolate the angle 2x, by following the reverse "order of operations".

Explanation:

Step 1: Add 1 to both sides:
`2cos^2(2x)=1`

Step 2: Divide both sides by 2:
`cos^2(2x) = 1/2`

Step 3: Take the square root of both sides:
`cos(2x) =(sqrt(2))/2 or cos(2x) =(-sqrt(2))/2`
(don't forget the positive and negative solutions!)

Step 4: Use inverse of cosine to find the angles:
`2x = cos^-1(sqrt(2)/2) or2x = cos^-1(-sqrt(2)/2)`

Step 5: Find angles that work:
`2x = pi/4 or 2x = (7pi)/4 or 2x=(3pi)/4 or 2x = (5pi)/4`

Step 6: Solve for x:
`x = pi/8, (7pi)/8, (3pi)/8, (5pi)/8` or .785, 5.5, 2.36, 3.93
(decimal approximations are seen on the graph below)