Question #84672

2 Answers
José F.
Dec 6, 2017

#x=kpi, k in ZZ#

Explanation:

#cos3x*cosx-1=0#

From tables take the formula for cos(3x):

#(4Cos^3x-3cosx)*cosx-1=0#

#4Cos^4x-3cos^2x-1=0#

This can be transformed in a second degree equation where #y=cos^2(x)#:

#4y^2-3y-1#

#y=(3+-sqrt(3^2-4*4*(-1)))/(2*4)#

#y=(3+-sqrt(9+16))/(8)=(3+-sqrt(25))/(8)#

#y=(3+-5)/(8)#

#y=8/8=1 or y=-2/8=-1/4#

Now replace y by Cox^2x:

#cos^2x=1 or cos^2x=-1/4#

#cosx=+-1 or impossible

#x=kpi #

Nghi N
Dec 6, 2017

#x = kpi#

Explanation:

Use trig identity:
#cos a.cos b = (1/2)[cos (a + b) + cos (a - b)]#
In this case:
#cos 3x.cos x = (cos 4x + cos 2x)/2 - 1 = 0#
cos 4x + cos 2x - 2 = 0
Replace in the equation cos 4x by #(2cos^2 2x - 1)# -->
#2cos^2 2x + cos 2x - 3 = 0#
Solve this quadratic equation for cos 2x.
Since a + b + c = 0, use shortcut. The 2 real roots are:
#cos 2x = 1#, and #cos 2x = c/(a) = - 3/2# (rejected)
cos 2x = 1. Unit circle -->
#2x = 0 + 2kpi# --> #x = kpi#
#2x = 2pi + 2kpi # --> #x = pi + kpi General answer #x = kpi# Check. #x = pi# --> #cos 3x = cos pi = - 1# --> #cos x = cos pi = -1#. #cos 3x.cos x - 1 = (-1)(-1) - 1 = 0#. Proved

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