math
posted by Lilly .
Solve the equation for 0<x<2π
√(2) sin(x)+√(2) cos(x)>0

divide by √2
sinx + cosx > 0
sinx > cosx
sinx/cosx > 1
tanx > 1
from the graph of the standard tangent curve of y = sinx and y = cosx
in the domain 0 < x < 2π we see that sinx is above cosx for
0 < x < 3π/4 OR 7π/4 < x < 2π 
revised solution:
divide by √2
sinx + cosx > 0
sinx > cosx
from the graph of the standard curve of y = sinx and y = cosx
in the domain 0 < x < 2π we see that sinx is above cosx for
0 < x < 3π/4 OR 7π/4 < x < 2π
I originally went with tanx, as seen in my first attempt,
obtained by dividing both sides by cosx
However, since division by cosx would result in worrying about positive and negatives divisors giving me reversals of the inequality sign, I just went with the second version of my solution 
or, recognize that what you have is
2sin(x+pi/4) > 0
so, x+pi/4 must be in QI or QII
pi/4 < x < 3pi/4
But, we want x>0, so add 2pi to make it positive:
pi/4 < x is the same as 7pi/4 < x < 2pi
answer is as shown above.
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