Question

cos(3x)=-1

Part I: Solve this equation for 3x, finding all values on the interval [0,2π] that satisfy the equation.

Part II: Use algebra to find all values of x between 0 and 2π that satisfy this equation.

Answer 1

Where does the cosine take on the value -1?

Answer 2

pi?

Answer 3

You are nearly done with this problem.

Answer 4

Hmm, so how would I progress from there?

Answer 5

nvm, it can be -1 and 1, but not more

Answer 6

You have a solution, 3x = pi.
You know the cosine is periodic with period 2pi

\(3x = \pi + 2k\pi\) for any integer \(k\).

Divide by 3.
Figure out what 'k' means.

Answer 7

Hmm, I just can't figure out what 'k' could mean, could you give me a bit of a hint? x)

Answer 8

Did already. It is any integer. Your task is to find which integers put the solution in \([0,2\pi]\).

Answer 9

Hm would it be something like -1?

Answer 10

?? You have the equation already: \(3x = \pi + 2k\pi\)

Solve for x. \(x = \dfrac{\pi}{3} + \dfrac{2k\pi}{3} = \dfrac{\pi}{3}\left(1 + 2k\right)\)

Solve for k

\(1+2k = 0\)
\(1+2k = 1\)
\(1+2k = 2\)
\(1+2k = 3\)
\(1+2k = 4\)
\(1+2k = 5\)
\(1+2k = 6\)

Why did I quit at 6?