Solving Trig Equation?

Question

cos4x(cosx-1)=0

Answer 1

cos(4x) [ cos(x) - 1 ] = 0

I'm going to presume you want a solution with the restriction
0 <= x < 2pi.

Equate each factor to 0.

cos(4x) = 0
cos(x) - 1 = 0

cos(4x) = 0
cos(x) = 1

Cosine is equal to 0 at the points pi/2 and 3pi/2. Since we have 4x, we're going to have quadruple times the normal solutions (i.e. 8 solutions instead of 2) for the first equation. We obtain these other solutions by adding 2pi, or 4pi/2, to each original solution for cos(4x).

4x = {pi/2, 3pi/2, 5pi/2, 7pi/2, 9pi/2, 11pi/2, 13pi/2, 15pi/2}

Multiply by (1/4) both sides,

x = {pi/8, 3pi/8, 5pi/8, 7pi/8, 9pi/8, 11pi/8, 13pi/8, 15pi/8 }

For cos(x) = 1, this is true when x = 0. Therefore our complete set of solutions in that restricted interval is

x = {pi/8, 3pi/8, 5pi/8, 7pi/8, 9pi/8, 11pi/8, 13pi/8, 15pi/8, 0}

Answer 2

Solve for x.

(cos4x)(cosx - 1) = 0

cos 4x = 0
4x = arccos(0) = π/2 + πk, where k is an integer
x = π/8 + (π/4)k, where k is an integer

cos x = 1
x = arccos(1) = 0 + 2πk, where k is an integer

So the complete solution is:

x = π/8 + (π/4)k and 0 + 2πk, where k is an integer