Question

The value of x for the minimum value of √3 cos +sin is *
(i) 180 degree
ii 270 degree
iii 210 degree
iv 0 degree​

Answer 1

Given,∣3cosx−sinx∣≥2 

⇒3cosx−sinx=±2

Put 3=rcosα,1=rsinα

⇒r=2,α=6π

So, the equation becomes 2cos(x+6π)=±2

⇒cos(x+6π)=±1

⇒x=65π,611π,617π,623π for  x∈[0,4π] 

Hence, there are four solutions.

Answer 2

Answer:

the upper one part is correct ans