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 
sec(θ)= 2 can be written as cos(θ)=1/2
Using unit circle the value of θ is 
For general solution we add 2npi
So critical points are
Answer:
Step-by-step explanation:
Knowing the three zeros of the equation, we can set it up as follows:
Multipying the first two expressions together gives us the following:
Multiplying the two expressions together gives us the following:
Answer:
The Answer is 2.
Step-by-step explanation:
Answer:
If the absolute value expression is not equal to zero, the expression inside an absolute value can be either positive or negative. So, there can be at most two solutions. Looking at this graphically, an absolute value graph can intersect a horizontal line at most two times.
Answer:
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