cosx = sqrt(2)/2?

Question

Please find all real numbers that satisfy this equation. Thanks in advance!

Please show your work so I can understand how to complete the problem.

Thank you all so much....I understand now!

Answer 1

x = ± π/4 + 2kπ , k is any integer

Edit: you need to find real numbers... that means the answers must be given in radians.... not in degrees... or revolutions...

you need to be familiar with the unit circle...
another way to recognize this is if you have an isosceles right triangle...
if the legs are A units, the hypotenuse is (√2)A units
the hypotenuse will be 90° or π/2
... while the other angles must be 45° or π/4

cosine is the adjacent side over hypotenuse... thus
cos π/4 = A/(√2)A = √2 / 2

meanwhile... the x-values would be the same on the first and fourth quadrant.. (that is where the negative came about...)

finally, all trigonometric functions are at least periodic with a period of 2π

that is why, the answer is
x = ± π/4 + 2kπ , k is any integer

§

Answer 2

Sqrt 2 2

Answer 3

You need to know a little geometry. Namely that in a 45 degree right triangle with hypotenuse 1, the two legs are eqch sqrt(2)/2. Now draw the terminal side of an angle in standard position on a unit circle with reference angle 45 degrees in the first and 4th quadrants. Draw perpendiculars from the point on the unit circle to the x-axis and label the sides. Inquad I they are both x=sqrt(2)/2 and r=1.

Since cosx = x-coordinate/r for a point on the terminal side where r=1, All the angles with these two sides as their terminal sides give an x such that cosx = sqrt(2)/2.

They are x =pi/4 + 2kpi and x = 7pi/4 + 2kpi

Answer 4

This is a 45 degree angle
The cosine is positive in Quadrants I and IV
Hence the answer is :
x = 45 +/-360n degrees and x = 315 +/-360n degrees, where
n = any real integer.

Answer 5

x = 45° or −45°... or any ±360° multiple you want to tack on.

x = arccos (½√2)