SOLUTION: Solve the equation for the interval [0,2л): (sin2x+cos2x)²=1 PLEASE HELP ME!!!!!

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Question 181754: Solve the equation for the interval [0,2л):
(sin2x+cos2x)²=1
PLEASE HELP ME!!!!!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
%28sin2x+%2B+cos2x%29%5E2+=+1
First, you need whatever is in the parentheses to be either
1 or -1, since 1%5E2+=+1
and %28-1%29%5E2+=+1
If I chart the equation
-----------------------
x --- 2x --- (sin2x + cos2x)
-----------------------------
0 --- 0 --- 1
pi%2F4 --- pi%2F2 --- 1
pi%2F2 --- pi --- -1
3%2Api%2F4 --- 3%2Api%2F2 --- -1
pi --- 2%2Api --- 1
5%2Api%2F4 --- 5%2Api%2F2 --- 1
3%2Api%2F2 --- 3%2Api --- -1
7%2Api%2F4 --- 7%2Api%2F2 --- -1
2%2Api --- 4%2Api --- 1
------------------------------
It looks like there are 9 values of x
that solve the equation
Another way to write the solution is:
n%2Api%2F4 for n= 0 to 8