Given the function f(x) =2sec(`theta)`+tan(`theta)` in the interval 0<theta<2 pi
We have to find the critical points.
So let us first take the derivative of the function and evaluate it to zero.
`f'(theta)=2sec(theta)tan(theta)+sec^2(theta)=0`
`(2sin(theta))/(cos^2(theta))+1/(cos^2(theta))``=0`
`2sin(theta)+1=0`
`sin(theta)=-1/2```
implies, theta= 7pi/6 and 11pi/6 (that comes in the interval 0<theta<2 pi)
These are...
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Given the function f(x) =2sec(`theta)`+tan(`theta)` in the interval 0<theta<2 pi
We have to find the critical points.
So let us first take the derivative of the function and evaluate it to zero.
`f'(theta)=2sec(theta)tan(theta)+sec^2(theta)=0`
`(2sin(theta))/(cos^2(theta))+1/(cos^2(theta))``=0`
`2sin(theta)+1=0`
`sin(theta)=-1/2```
implies, theta= 7pi/6 and 11pi/6 (that comes in the interval 0<theta<2 pi)
These are the critical points.
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