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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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}.

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