Problem 9 is correct. Nice work.
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Problem 10 however, is incorrect.
Keep in mind that,
tan(x) = sin(x)/cos(x)
If we tried to plug in x = pi/2, then we'd get an error.
Why? Because cos(x) is equal to 0 when x = pi/2
The denominator becomes 0 and we can't divide by zero.
Tan(x) is undefined when x = pi/2
So overall, x = pi/2 is not part of the domain of tan(x).
Consequently, x = pi/2 is not part of the domain of f(x) = cos(x)-tan(x)*cos(x)
This allows us to rule out choice C and choice D
The answer comes down to either A or B. Let's try choice A.
Plug in x = 0 and see what happens.
f(x) = cos(x)-tan(x)*cos(x)
f(0) = cos(0)-tan(0)*cos(0)
f(0) = 1 - 0*1
f(0) = 1
We don't get a result of 0 like we want, so x = 0 is not a root of f(x)
We can cross choice A off the list.
Let's see what happens if we try out the x values for choice B
Plug in x = pi/4
f(x) = cos(x)-tan(x)*cos(x)
f(pi/4) = cos(pi/4)-tan(pi/4)*cos(pi/4)
f(pi/4) = sqrt(2)/2-1*sqrt(2)/2
f(pi/4) = 0
Now plug in x = 5pi/4
f(x) = cos(x)-tan(x)*cos(x)
f(5pi/4) = cos(5pi/4)-tan(5pi/4)*cos(5pi/4)
f(5pi/4) = -sqrt(2)/2-1*(-sqrt(2)/2)
f(5pi/4) = 0
Both inputs x = pi/4 and x = 5pi/4 lead to f(x) = 0
So these inputs are roots of f(x) on the interval [0,2pi]
The answer for problem 10 is choice B
