cosx/2-cosx=0?!?!?!?

Question

Anyone willing to show me how to do this one?Thanks!

Answer 1

cos x/(2-cos x) = 0
means cos x = 0
x = pi/4

in case you meant cos x/2- cos x = 0

then let cos x/2 = t

cos x = 2t^2-1

so t - (2t^1-1) = 0
or 2t^2 -t + 1 = 0

we can solve for t and get x

Answer 2

The cosines of 2 arcs are equal if, and only if, they have the same determination or if their determination are symmetric with respect the horizontal axis.

In the first case we must have
x/2 = 2kpi + x for integer k. Hence, x = 4kpi

In the second case, we must have x/2 = 2kpi - x, hence
x = 4kpi/3

So, the solotion are the arcs of the form x = 4kpi and x = 4kpi/3.

Answer 3

cos x = cos (x/2 + x/2) = cos²(x/2) - sin²(x/2) = 2 cos²(x/2) -1

Then your equation becomes

2 u² - u -1 = 0,

where u = cos(x/2)

Solving for u, we obtain

u = [1 ± √(1+8)] / 4 = [1 ± 3] / 4

u1 = 1, u2 = -½

cos(x/2) = 1; then x/2 = 2 k π, or x = 4 k π.

cos(x/2) = -1/2, then x/2 = 2 k π ± 2 π / 3, or
x = 4 (k ± 1/3) π