Question

Solve: 2 cos^2 x - 3 cos x + 1 = 0 for 0< x<2pi
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Answer 1

How would you rewrite this equation if b=cos(x)? and you subbed b for every instance of cos(x)?

Answer 2

idk

Answer 3

\[2\cos(x)\cos(x) -3\cos(x) +1 =0\]
so what does this equation become when you replace every instance of cos(x) with the letter b?

Answer 4

Do you still need help with this question?

Answer 5

Please reply so I would know If i should continue.

Answer 6

you can continue

Answer 7

@zordoloom

Answer 8

Okay, sorry I was helping someone else. So Start by identifying what this equation really is
...

Answer 9

This is an ax^2+bx+c problem

Answer 10

Which means that it might be possible to factor so....

2 cos^2 x - 3 cos x + 1 = 0 ------>

(cosx-1/2)(cosx-1)=0 is the factored form

Let me know if I should continue.

Answer 11

yes..

Answer 12

Now, this also means that the factors can be rewritten as...

(2cosx-1)(cosx-1)=0

Now solve the first equation

(2cosx-1)=0------>

2cosx-1=0

Add 1,

2cosx=1

Divide by 2

cosx=1/2

Answer 13

Now take the arccos(1/2). There is only solution from 0 to 2pi. Which is --->

pi/3

Now solve the second equation.

Answer 14

Actually, sorry, there would be 2 solution for cosx=1/2

p/3, and 5pi/3

Now, solve the second equation.

Answer 15

Do you follow so far?

Answer 16

Continue?

Answer 17

(cosx-1), now solve this for x,

Add 1

cosx=1

Take the arccos(1)

You get 2pi