Question

Question #84672

Answer 1

`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`

Answer 2

`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