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

3 Answers
Sunayan S
Dec 4, 2017

See the answer below...

Explanation:

We know that , " " color(red)(sin^2x+cos^2x=1sin2x+cos2x=1

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

sin^2x+cos^2x=1sin2x+cos2x=1
=>(sinx+cosx)^2-2cdot sinx cdot cosx=1⇒(sinx+cosx)2−2⋅sinx⋅cosx=1
=>(sinx+cosx)^2=1+2 cdot sinx cdot cosx⇒(sinx+cosx)2=1+2⋅sinx⋅cosx
[The value of color(red)(2 cdot sinx cdot cosx2⋅sinx⋅cosx might be color(red)(0)0 but color(red)(1>01>0]
=>(sinx+cosx)^2>0⇒(sinx+cosx)2>0
=>sinx +cosx>0⇒sinx+cosx>0

Hope it helps...
Thank you...

Cesareo R.
Dec 4, 2017

pi/4 < x < (5pi)/4π4<x<5π4

Explanation:

sinx+cosx = sinx+sin(pi/2-x) sinx+cosx=sinx+sin(π2−x)

and

sin(a+b)+sin(a-b) = 2sina cosbsin(a+b)+sin(a−b)=2sinacosb

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

Nghi N
Dec 5, 2017

(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)

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