Answer:
[tex]\theta = n\pi[/tex]
where [tex]n = ..., -2,-1,0,1,2...[/tex]
Step-by-step explanation:
given equation is:
[tex]\cos{2\theta} - 1 = 0[/tex]
since no range is provided we can solve for all values of [tex]\theta[/tex]:
[tex]\cos{2\theta} = 1[/tex]
[tex]2\theta = \cos^{-1}{(1)}[/tex]
[tex]2\theta = 0, 2\pi[/tex] for one cycle of cos [tex](0 \leq \theta \leq 2\pi)[/tex]
[tex]2\theta = 0, 2\pi, 4\pi, 6\pi ... 2n\pi[/tex] for all cycles of cos
we should also include negative values.
[tex]2\theta = -4\pi,-2\pi,0, 2\pi, 4\pi,... 2n\pi[/tex]
we can divide each value by 2, to get the solutions for [tex]\theta[/tex] instead of [tex]2\theta[/tex]
Answer:
[tex]\theta = -2\pi,-\pi,0, \pi, 2\pi,... n\pi[/tex]
This is the solution of the equation [tex]\cos{2\theta} - 1 = 0[/tex].
In its most general form we can write all solutions of the equation in terms of [tex]n[/tex]
[tex]\theta = n\pi[/tex] where [tex]n = ..., -2,-1,0,1,2...[/tex] or n is an integer.