SOLUTION: Solve the following for 0 < x < 2pi (sin x + cos x)^2 = 1

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Question 937255: Solve the following for 0 < x < 2pi
(sin x + cos x)^2 = 1
Found 2 solutions by lwsshak3, Alan3354:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the following for 0 < x < 2pi
(sin x + cos x)^2 = 1
(sinx+cosx)^2=sin^2(x)+2sinxcosx+cos^2(x)=1+sin(2x)
1+sin(2x)=1
sin2x=0
2x=0
x=0
no solution: x not in given domain

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the following for 0 < x < 2pi
(sin x + cos x)^2 = 1
sin(x) + cos(x) = 1
By inspection, x = pi/2
------
sin(x) + cos(x) = -1
x = pi