How to solve #cos(x)+sin(x) > 0#?
3 Answers
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...
Explanation:
and
making now
we get
Explanation:
sin x + cos x > 0
Use trig identity:
On the unit circle, cos (x - pi/4) > 0 --> 2 solutions:
a. the arc
b. The arc
Answers by intervals: