Find the angle a if 2cos^2a+1=3cosa

The first step is to move all terms to the left
side:

2(cos a)^2- 3cos a + 1 =
0

Now, we'll use substitution technique to solve the
equation.

We'll note cos a = 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 a = t1.

cos a= 1

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

cos
a = y

a = +/- arccos y +
2k*pi

In our case, y = 1:

a
=  +/-arccos 1 + 2k*pi

a = 0 +
2k*pi

a = 2k*pi

Now, we'll
put cos a = t2

cos a = 1/2

a =
+/- arccos 1/2 + 2k*pi

a = +/- (pi/3) +
2k*pi

a = 2k*pi + pi/3

a =
2k*pi - pi/3

The solutions of the equation
are:{  2k*pi}U{2k*pi - pi/3}U{2k*pi + pi/3}.