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)?

1 Answer
A. S. Adikesavan
Aug 6, 2018

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 .

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