Question

How to solve `cos(x)+sin(x) > 0`?

Answer 1

See the answer below...

Explanation:

We know that , `" "` `color(red)(sin^2x+cos^2x=1`

Let's use the inequality here.....

`sin^2x+cos^2x=1`
`=>(sinx+cosx)^2-2cdot sinx cdot cosx=1`
`=>(sinx+cosx)^2=1+2 cdot sinx cdot cosx`
[The value of `color(red)(2 cdot sinx cdot cosx` might be `color(red)(0)` but `color(red)(1>0`]
`=>(sinx+cosx)^2>0`
`=>sinx +cosx>0`

Hope it helps...
Thank you...

Answer 2

`pi/4 < x < (5pi)/4`

Explanation:

`sinx+cosx = sinx+sin(pi/2-x)`

and

`sin(a+b)+sin(a-b) = 2sina cosb`

making now

`{(a+b=x),(a-b=pi/2-x):}`

we get

`{(a = pi/4),(b=x-pi/4):}` and then

`sinx+cosx =2sqrt2/2cos(x-pi/4)` or

`cos(x-pi/4) > 0` and then

`-pi/4 < x < 3/4pi`

Answer 3

`(pi/4, (3pi)/4) and ((7pi)/4, pi/4)`

Explanation:

sin x + cos x > 0
Use trig identity:
`sin x + cos x = sqrt2cos (x - pi/4) > 0`
`cos (x - pi/4) > 0`
On the unit circle, cos (x - pi/4) > 0 --> 2 solutions:

a. the arc `(x - pi/4)` lays between 0 and `pi/2`
`0 < (x - pi/4) < pi/2`
`pi/4 < x < pi/2 + pi/4 = (3pi)/4`

b. The arc `(x - pi/4)` lays between `(3pi)/2` and `2pi` -->
`(3pi)/2 < (x - pi/4) < 2pi`
`(7pi)/4 < x < pi/4`
Answers by intervals:
`(pi/4, (3pi)/4)`, and `((7pi)/4, pi/4)`