Question #899b3

1 Answer
Jan 3, 2018

#x = (5pi)/12#

Explanation:

#sinx + cosx = 1/sqrt2#

Squaring both sides,
#sin^2x + cos^2x + 2sinxcosx = 1/2#

As #sin^2x + cos^2x = 1#
So,
#2sinxcosx = -1/2#

As #sin2x = 2sinxcosx#

#sin2x = -1/2#

#sin2x = sin((5pi)/6)#

#2x = (5pi)/6#

#x = (5pi)/12#

......................