What are the values of #x# that satisfy #cos^2x = 1/2#?

1 Answer
Apr 20, 2017

Solution is >#x in {45˚ + 360˚n, 135˚ + 360˚n, 225˚ + 360˚n, 315˚ + 360˚n}# where #n# is an integer.

Explanation:

Here's another way of doing this problem.

Take the square root of both sides.

#cosx = +- sqrt(1/2)#

#cosx = +- 1/sqrt(2)#

Now consider the #45-45-90# special triangle.

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We have:

#x = 45˚#

But this is just the reference angle for the three other solutions in #0 ≤ x < 360˚#. There will also be #x = 135˚#, #x = 225˚# and #x = 315˚#.

If you want the periodic solutions, we have:

#x in {45˚ + 360˚n, 135˚ + 360˚n, 225˚ + 360˚n, 315˚ + 360˚n}# where #n# is an integer.

Hopefully this helps!