Question

find 2 values of x for where f(x)=0. f(x) = −2 cos(2x + π) + 1

Answer 1

You are being asked to solve -2cos(2x+pi)+1=0

Answer 2

isolate the trig function part.

Answer 3

Then use the handy dandy unit circle.

Answer 4

what makes it so handy dandy

Answer 5

I don't understand... so like cos(2x+pi) = 0?

Answer 6

-2cos(2x+pi)+1=0
add +2cos(2x+pi) to both sides
0 + 1 = 2cos(2x+pi)
2cos(2x+pi) = 1
divide both sides by 2
cos(2x+pi) = 1/2
Find two values when cosine is 1/2. Substitute and solve for x.

Answer 7

2 values when cosine is 1/2 where? If I graph it would it help?

Answer 8

do you mean on the unit circle? because then I believe that is pi/3 and 5pi/3

Answer 9

calculator will tell you. It is called arc cos or inverse cosine marked with cos^-1.
Or you can look up trig tables.

Answer 10

yes pi/3 and 5pi/3 are good.
2x + pi = pi/3. solve for x.

2x + pi = 5pi/3. solve for x.

Answer 11

what about a value for x when f(x) is minimized? any clue what that means?

Answer 12

cosine function goes from +1 to -1.
f(x) = −2 cos(2x + π) + 1
Put the max and min values for cosine and find f(x)
f(x) = -2(1) + 1 = -1
f(x) = -2(-1) + 1 = 3

So f(x) has a minimum when the cosine function is maximum at 1.
Solve for x when cos(2x + pi) = 1

Answer 13

any suggestions about finding the range of f(x)?

Answer 14

and is the answer to the minimizing just 2x+pi=2pi and solve? because thats the only time cosine = 1, right?