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)#?

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Answer 1 · 2018-08-06T11:51:52

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

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 #.