What is x for 2cos^2x - 3cosx + 1 = 0 ?

It is obviously that we'll use substitution technique to
solve the equation.

We'll note cos x = t and we'll re-write
the equation in t:

2t^2 - 3t + 1 =
0

Since it is a quadratic, we'll apply the quadratic
formula:

t1 = {-(-3) + sqrt[(-3)^2 -
4*2*1]}/2*2

t1 =
[3+sqrt(9-8)]/4

t1 =
(3+1)/4

t1 = 1

t2 =
(3-1)/4

t2 = 1/2

Now, we'll
put cos x = t1.

cos x =
1

Since it is an elementary equation, we'll apply the
formula:

cos x = a

x = +/-
arccos a + 2k*pi

In our case, a =
1:

x = +/- arccos 1 + 2k*pi

x
= 0 + 2k*pi

x =
0

or

x =
2pi

Now, we'll put cos x =
t2

cos x = 1/2

x = +/-arccos
(1/2) + 2k*pi

x = +/- pi/3 +
2k*pi

x = pi/3

x = pi -
pi/3

x =
2pi/3

The solutions of the equation are:{0 ;
pi/3 ; 2pi/3 ; 2pi}.