Question

What are the critical points for g(θ) = 20θ − 5 tan θ

Answer 1

g(θ) = 20θ − 5 tan θ

To find out critical points we take first derivative and set it =0

g(θ) = 20θ − 5 tan θ

g'(θ) = 20 − 5 sec^2(θ)

Now we set derivative =0

20 − 5 sec^2(θ)=0

Subtract 20 from both sides

− 5 sec^2(θ)=0 -20

Divide both sides by 5

sec^2(θ)= 4

Take square root on both sides

sec(θ)= -2  and sec(θ)= +2

sec can be written as 1/cos

so sec(θ)= -2  can be written as cos(θ)= -1/2

Using unit circle the value of θ is $\frac{2\pi}{3} and\frac{4\pi}{3}$

sec(θ)= 2  can be written as cos(θ)=1/2

Using unit circle the value of θ is $\frac{\pi}{3}and\frac{5\pi}{3}$

For general solution we add 2npi

So critical points are

$\frac{2\pi}{3} + 2n\pi,\frac{4\pi}{3} + 2n\pi , \frac{\pi}{3} + 2n\pi, \frac{5\pi}{3} + 2n\pi$