SOLUTION: How to: "Find the number of solutions of the equation cos^2-1=0 in the interval [0, 2pi] I began by using the Pythagorean identity and replaced cos^2 by (1-sin^2x). The proble

Algebra ->  Trigonometry-basics -> SOLUTION: How to: "Find the number of solutions of the equation cos^2-1=0 in the interval [0, 2pi] I began by using the Pythagorean identity and replaced cos^2 by (1-sin^2x). The proble      Log On


   

Question 888438: How to: "Find the number of solutions of the equation cos^2-1=0 in the interval [0, 2pi]
I began by using the Pythagorean identity and replaced cos^2 by (1-sin^2x).
The problem then looked like: (1-sin^2x)-1=0.
I canceled out the 1 & -1 making the problem: -sin^2=0
I do not know where to go from here, or if I have been doing the problem right so far.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
You're alright up to there although you could have just stayed with cosine.
-sin%5E2%28x%29=0
Multiply both sides by (-1).
sin%5E2%28x%29=0
Take the square root of both sides.
sin%28x%29=0
Now, which values of x give you a value of 0?
There are two of them.
x=0
and
x=pi
.
.
.
For the cosine,
cos%5E2%28x%29-1=0
cos%5E2%28x%29=1
cos%28x%29=0+%2B-+1
Also, two solutions.
x=0 and
x=pi