Find x.Find x for 2cos^2x - 3cosx + 1 = 0.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to solve for x the trigonometric equation using factorization, such that:

`2cos^2x - 3cosx + 1 = 0 => 2cos^2x - (2 + 1)cosx + 1 = 0`

`2cos^2 x - 2cos x - cos x + 1 = 0`

You need to group the terms such that:

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

`2cos x(cos x - 1) - (cos x - 1) = 0`

Factoring out `(cos x - 1)` yields:

`(cos x - 1)(2cos x - 1) = 0 => {(cos x - 1 = 0),(2cos x - 1 = 0):}`

`{(cos x = 1),(2cos x = 1):} => {(x = +-cos^(-1) 1 + 2npi),(cos x = 1/2):} => {(x = 2npi),(x = +-pi/3 + 2npi):}`

Hence, evaluating solutions to trigonometric equation yields` x = 2npi, x = +-pi/3 + 2npi.`

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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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