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

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Answer 1 · 2016-02-16T08:20:41

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

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 · 2016-02-16T08:32:21

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

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..