Calculus of a Single Variable, Chapter 3, 3.1, Section 3.1, Problem 15
Given the function h(x)=sin^2(x)+cos(x) in the interval 0
First take the derivative of the function and equate it to zero.
We get,
h'(x)=2sin(x)cos(x)-sin(x)=0
sin(x)(2cos(x)-1)=0
sin(x)=0 or 2cos(x)-1=0
sin(x)=0 implies x= npi
i.e we get x= pi in the interval 0
2cos(x)-1=0 implies cos(x)=1/2
So x= pi/3 and 5pi/3 (in the interval 0
Comments
Post a Comment