Are there any solutions for this equation within (0,2π)? [cosx=sinx] if so, what are they?

1 Answer
Apr 18, 2018

#x=pi/4, (5pi)/4#

Explanation:

Examining the unit circle, we see where #cosx=sinx#, or where #(x,y)=(x,x)#:

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#x=pi/4, (5pi)/4#

It does not hold true in the other quadrants (quadrants II, IV), as the sine and cosine have opposite signs for all angles in those quadrants.