Solve for x : 3cosx - 1 = 2cos^2x

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changchengliang's profile pic

changchengliang | Elementary School Teacher | (Level 2) Adjunct Educator

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Re-arranging the equation,

2(cosx)^2 - 3cosx + 1 = 0

Factorising,

(2cosx-1) (cosx-1) = 0

cosx = 1/2   or  1

Hence x can take on 360*n deg, (360*n+60)deg, (360*n-60)deg

 

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to solve 3*cos x - 1 = 2*(cos x)^2

First let's bring all terms to one side.

3*cos x - 1 = 2*(cos x)^2

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

=> 2*(cos x)^2 - 2*cos x - cos x + 1 = 0

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

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

For 2 cos x -1 =0 , we have cos x = 1/2

and for cos x -1 = 0, we have cos x =1

If cos x = 1/2, x = arccos ( 1/2)= 60 degrees and 360-60 = 300 degrees.

If cos x = 1, x = arccos (1) = 0 degree.

Therefore the values of x are 0 + 360*n degree, 60 + 360*n degrees and 300 + 360*n degrees.

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neela | High School Teacher | (Level 3) Valedictorian

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3cosx -1 = 2cos^2x.

The above is a quadratic  cosx . So we arrange the equation like as below by subtracting 3cosx-1 from both sides:

2cos^2x - 3cosx +1 = o

2c^2-3c +1 = 0, where c= cosx.

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 gives cosx = 1, or x = 0 or 2pi.

2x= 1gives 2cosx = 1. Or cosx = 1/2. Or  x = pi/3 or x = 60 deg, Or x = 2pi-pi/3 = 4pi/3 , or x = 300 deg.

Solution: x= 0 , pi/3 , 4pi/3 or 0 deg, 60 deg or 300 deg.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The first step is to move all terms to one side:

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

We'll solve the equation using substitution technique.

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 = (-1)^k* arccos a + 2k*pi

In our case, a = 1:

x =  (-1)^k* arccos 1 + 2k*pi

x = (-1)^k*(0) + 2k*pi

x = 2k*pi or x = 2pi + 2k*pi

x = 0

x = 2pi(k+1)

x = 2pi

Now, we'll put cos x = t2

cos x = 1/2

x = (-1)^k* arccos 1/2 + 2k*pi

x = (-1)^k* (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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