2cos^2 x - cosx = 0... Help solving?

Question

Find the exact answers:

2cos^2 x - cosx = 0
0 <= x < 2(pi)

That first "x" is not connected to the exponent (it's just the cos variable). I don't know what to do! An explanation would be just lovely.

Answer 1

Factoring out cos(x) gives
cos(x) (2cos(x) - 1) = 0

So cos(x) = 0, or 2cos(x) - 1 = 0. Which means cos(x) = 0 or cos(x) = 1/2. Write down the values of x that satisfy either of these conditions. You should get four.

Answer 2

You can take the cosx function and substitute it for another variable, say y, and try to solve that first.

So the equation would become:
2y^2 - y = 0
y (2y - 1) = 0
y = 0, 1/2

That means that:
cosx = 0, and cosx = 1/2

For the interval 0 <= x < 2pi, the values that fit for the first equation are:
cosx = 0
x = pi/2, 3pi/2

For the second equation, the angles that fit are:
cosx = 1/2
x = pi/3, 5pi/3

So the values of x are:
x = pi/3, pi/2, 3pi/2, 5pi/3

Answer 3

2cos^2(x) - cosx = 0
cosx(2cosx - 1) = 0

now we get two possibilities,

cosx = 0 ; cosx = 1/2
x = cos^(-1) 0 ; x = cos^(-1) 1/2
x = pi/2, 3pi/2 ; x = pi/3, 5pi/3

Answer 4

(cos x) (2cos x - 1) = 0
cos x = 0 , cos x = 1 / 2
x = π / 2 , 3π / 2 , π / 3 , 5π / 3

Answer 5

cos[x] + cos{x} = 2 cos([x]+{x} / 2) cos([x]-{x} / 2) = 2 cos x cos ([x]-{x} / 2) 2 cos x cos ([x]-{x} / 2) = sin x =>cos ([x]-{x} / 2) = 1/2 tan x Only one solution at x = 45