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

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To find angle a , if 2cos^2a +1 = 3cosa.

This is a quadratic equation in cosa. So we solve for cosa and then determine angle a.

We put cosa = c in the given equation:

2c^2+1 = 3c.

We subtract 3c from both sides:

2c^2 -3c +1 = 0

2c^2-2c -c +1 = 0.

2c(c-1) -1(c-1) = 0

(c-1)(2c-1) = 0.

c-1 = 0 , or 2c-1 = 0.

c = 1, or c = 1/2

c= 1 gives cosa = 1, or a = 0 or a = 2npi.

c = 1/2 gives cosa = 1/2. a = pi/3. Or a = 2npi+pi/3. or 2npi-pi/3

Therefore , a = 2npi, or 2npi + or - pi/3, n = 0,1,2...

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}.**