Question

Find all values of t in the interval [0, 2 π] that satisfy cos^2 t-4cost+3=0 . (Hint: Substitute x = cos t, factor the result, find the values of x that solve the equation, then put cos t back in for x.)

Answer 1

Comment: The equation should be cos2 t - 4cos t + 3 = 0. Rewrite this as x2 - 4x + 3 = 0 and factor to solve for x. Now going backward cos t = the numbers you got for x. You should be able to find the correct answers from the Unit Circle.

Answer 2

I solved x2 - 4x + 3 = 0and got x=1 and x=3

Answer 3

good work,
now,
\[\cos t=1\qquad\qquad\qquad\cos t=3\]

Answer 4

for what values of "t" is cos(t)=1??

Answer 5

I got 2piN1

Answer 6

I do not understand that

Answer 7

When it says backwards, is it wanting the cos^-1?

Answer 8

yep

Answer 9

your domain is a closed interval..
so
when t = 0 and
when t = \(2\pi\)
\(\cos(t)=1\)
correct?

Answer 10

I got 0.5403.

Answer 11

how is that?

Answer 12

I did cos(1)

Answer 13

no
sine \(\cos(t)=1\)
we do
\(t=\cos^{-1}(1)\)
get it?

Answer 14

now read my previous setps

Answer 15

I got t=0.

Answer 16

can there be any other possibilities?

Answer 17

2pi?

Answer 18

There is no solution for t=cos^-1(3)

Answer 19

good.
all other values will not fit in our domain.
so, these are the two solutions for the first one
now,
the last one t=cos^-1(3)
can never happen

Answer 20

good job.

Answer 21

Thank you for the help!

Answer 22

yw