How do you find all solutions of 2cos^2(θ)-1=0 to be pi/4 +kpi and 3pi/4 + kpi? I keep getting 2kpi instead of kpi

2cos^2θ-1=0
2cos^2θ=1
cos^2θ=1/2
cosθ=1/sqrt2=sqrt2/2
θ=±pi/4 +2kpi

θ=pi/4 +2pik
or
θ=3pi/4 +2pik

but my book says the answer is:

θ=pi/4 +kpi
or
θ=3pi/4 +kpi

1 Answer
Oct 17, 2016

2cos^2(theta)-1 = 0

=> cos^2(theta) = 1/2#

=> cos(theta) = +-sqrt(1/2) = +-sqrt(2)/2

Checking our unit circle, we find that cos(theta) = sqrt(2)/2 at theta = pi/4+2pik and theta=(7pi)/4 + 2pik. For the negative value, we get cos(theta)=-sqrt(2)/2 for theta=(3pi)/4+2pik and theta=(5pi)/4+2pik.

The problem with your given answer is that it does not account for solutions at which the angle is equivalent to (5pi)/4 or (7pi)/4. Now let's see how the book was able to put it in the given form.

Noting that pi/4 and (5pi)/4 are offset by exactly pi, we know (5pi)/4+2pik will just be pi/4+2pik+pi = pi/4+pi(2k+1). Thus, we can combine those answers to be pi/4+pik, as adding pi an even number of times will result in us landing on pi/4 and adding pi an odd number of times will result in us landing on (5pi)/4.

Similarly, we can combine (3pi)/4+2pik and (7pi)/4+2pik into a single answer of (3pi)/4+pik, as an even number of rotations of pi will put us back on (3pi)/4 and an odd number will put us on (7pi)/4.

The book stops there, but if we wanted to put the answer in its most concise form, we could just note that every answer is an integer number of rotations of pi/2 from pi/4, and that any such rotation will land on a solution. Thus, the solution set could also be written as

x=pi/4+(pi/2)k