Question

How do you use the inverse functions where needed to find all solutions of the equation `2cos^2x-5cosx+2=0` in the interval `[0,2pi)`?

Answer 1

See explanation.
In [ 0, 2pi ], the solutions are
`x = 1/3pi and 5/3pi`.

Explanation:

Solving,

`cos x = ( -(- 5) +- sqrt ( 25 - 16 ))/4= ( 5 +- 3 )/4 = 1/2 and 2`

As `cos x in [ -1, 1 ]`, discard `2 > 1`. So,

`cos x = 1/2 = cos (pi/3)`

Inverting ,

`x = cos^(-1)(cos ( pi/3 ) = pi/3 in [ 0, pi ]`,

picked from the general solution

`x = (cos)^(-1) cos (pi/3) = 2kpi +- pi/3, k = 0, +-1, +-2, +-3, ..`

`= ...-13/3pi, -11/3pi, -7/3pi, -5/3pi, -1/3pi, (1/3pi, 5/3pi), 7/3pi,...`

`= ...+-1/3pi, +-5/3pi, +-7/3pi+-11/3pi+-13/3pi, ...`

I have been insisting on the use of the piecewise wholesome

inverse operator `( cos )^(-1)` along with the conventional

`cos^(-1)`

that dwells only in the restricted interval `[ 0, pi ]`.

In brief,

if `cos x = cos alpha`,

`x = cos^(-1) cos alpha = alpha`, if and only if `alpha in [ 0, pi ]`.

Otherwise, the one that befits the cosine value, from the interval

`[ 0, pi ]`, is displayed.

This algorithm is used in calculators.

For example,

`cos^(-1) cos ( -60^o)` results in `60^o`,

using `cos (-60^o) = cos 60^o`.