Solve |#sinx + cosx#| = |#sinx#| + |#cosx#|, x#in# [0, 2#pi#] ?

1 Answer
Apr 19, 2018

#x in [0,pi/2] cup [pi,(3pi)/2]#

Explanation:

For real numbers #x# and #y#, the relation

#|x+y| = |x|+|y|#

works as long as both #x# and #y# are both positive, or both negative.

This means that for

#|sin x+cos x| = |sin x|+|cos x|#

to hold :

  • either both #sin x# and #cos x# are positive - hence #x# is in the first quadrant.
  • or, both are negative - hence #x# is in the third quadrant.