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

3 Answers
Dec 4, 2017

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...

Dec 4, 2017

#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#

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)#