Question

How do you find all values of x in the interval [0, 2pi] in the equation `2 + cos2x = 3cosx`?

Answer 1

Solution set is `{0, pi/3, (5pi)/3, 2pi}`.

Explanation:

As `cos2x=2cos^2x-1`, the function `2+cos2x=3cosx` is equal to `2cos^2x-3cosx+1=0`.

or `2cos^2x-2cosx-cosx+1=0` or

`2cosx(cosx-1)-1*(cosx-1)=0` or

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

`cosx=1/2` or `cosx=1` i.e.

if `x` is in first quadrant, `x=pi/3` or `x=0`

But as cosine of an angle is also positive in 4th quadrant, another solution could be `x=2pi-pi/3=5pi/3` and `x=2pi`

Hence solution set is `{0, pi/3, (5pi)/3, 2pi}`.

Answer 2

`x=0^@,60^@,300^@,360^@` or
`x=0, pi/3, (5pi)/3,2pi" "`radians

Explanation:

from the given

`2+cos 2x=3 cos x`
`2+cos^2 x-sin^2 x-3 cos x=0`

Recall that `sin^2 x=1-cos^2 x`
so that the equation becomes

`2+cos^2 x-(1-cos^2 x)-3 cos x=0`

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

solution by factoring

`(2cos x-1)(cos x-1)=0`

Equate both factors to zero

`2cos x-1=0` and `cos x-1=0`

`x=cos^-1 (1/2)=60^@, 300^@`
`x=cos^-1 (1)=0^@, 360^@`

have a nice day ! from the Philippines..