Cos 2x = 2 cos x where x is between 0 to 360 degree in trigonometrics?

2 Answers
Oct 25, 2017

The solutions are #S={111.5^@,248.5^@}# in the interval #[0,360]#

Explanation:

We need

#cos(2x)=2cos^2x-1#

Therefore,

#cos2x=2cosx#

#2cos^2x-1=2cosx#

#2cos^2x-2cosx-1=0#

This is a quadratic equation in #cos^2x#

The discriminant is

#Delta=b^2-4ac=(-2)^2-4(2)(-1)=4+8=12#

As #Delta>0#, there are 2 real roots

#cosx=(-b+-sqrtDelta)/(2a)=(2+-sqrt12)/4#

#=(1+-sqrt3)/2#

#cosx=(1+sqrt3)/2=1.37#, there is no solution since #-1<=cosx<=1#

#cosx=(1-sqrt3)/2=-0.366#

#=>#, #x=111.5^@# and #x=248.5^@#

Oct 25, 2017

#x=111.47^circ# or #x=248.53^circ# (approximately)

Explanation:

By the double angle formula
#cos(2x)=2cos^2(x)-1#

So
#color(white)("XXX")cos(2x)=2cos(x)#
implies
#color(white)("XXX")2cos^2(x)-1=2cos(x)#
or
#color(white)("XXX")2cos^2(x)-2cos(x)-1=0#

Using the quadratic formula, this gives solutions at
#color(white)("XXX")cos(x)=(1+-sqrt(3))/2#
but
we know #cos(x) in [-1,+1]# and #(1+sqrt(3))/2 > +1#
so only
#color(white)("XXX")cos(x)=(1-sqrt(3))/2# is non-extraneous.

Using the #arccos# function gives us
#color(white)("XXX")x = 111.4707014^circ# for #x in [0,180^circ]#
(from here on I will approximate this as #111.47^circ#)

This is equivalent to a reference angle of #180^circ-111.47^circ=68.53^circ# relative to the negative X-axis
and
we know that (based on CAST) there will be another angle in the range #(180^circ,360^circ)# that, using this same reference angle, will give the same #cos# value, namely:
#color(white)("XXX")180^circ+68.53^circ=248.53^circ#