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

1 Answer
Mar 28, 2016

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)

my screen shot