Posted by **Lilly** on Monday, July 22, 2013 at 7:43pm.

Solve the equation for 0<x<2π

√(2) sin(x)+√(2) cos(x)>0

- math -
**Reiny**, Monday, July 22, 2013 at 8:03pm
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**

- use 2nd solution - math -
**Reiny**, Monday, July 22, 2013 at 8:09pm
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

- math -
**Steve**, Tuesday, July 23, 2013 at 5:18am
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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