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g(x)=1/sin x and f(x)= 1 / cos x find the value of x when the rate of g(x) = the rate of f(x) x in[0,2pi]

g(x)=1/sin x and  f(x)= 1 / cos x find the value of x when the rate of  g(x) = the rate of f(x)       x in[0,2pi]

Grade:12

1 Answers

vikash chandola
21 Points
10 years ago

Hello baha

The rate of change of a function means its derivative so when  you talk about rate of change of f(x) it is f ‘(x)  and the rate of change of g(x) is g ‘(x) according to question rate of change of both function is equal means f ‘(x) =g’(x)

f ’(x) = -cos(x)/((sinx)*(sinx)) and g’(x)=sin(x)/((cosx)*(cosx))

f ‘(x)=g’(x)

 

-cos(x)/((sinx)*(sinx))= sin(x) /((cosx)*(cosx))

-(cos(x)) ^3 = (sin(x))^3

This implies that

-cos(x)= sin(x)

Hence x belongs to (pi/2,pi)union(-pi/2,0)

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