# Solve: 2cos^2 - 7cosx + 3 = 0 Solve exactly for special angles, to 2 decimal places otherwise. Let cosx= t

Then,

2t^2-7t+3=0

The roots are

t1= (7+sqrt(7^2-4*2*3))/(2*2)

or

t2= (7-sqrt(7^2-4*2*3))/(2*2)

t1= (7+sqrt(25))/(4)

t1= (7+5)/(4)

t1= 3

t2= (7-sqrt(7^2-4*2*3))/(2*2)

t2= (7-sqrt(25))/(4)

t2= (7-5)/(4)

t2= 0.5

But -1<=cosx<=1 ; hence cosx cannot be 3.

Then cosx= 0.5

The primary solution is x= pi/3

The general solution...

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Let cosx= t

Then,

2t^2-7t+3=0

The roots are

t1= (7+sqrt(7^2-4*2*3))/(2*2)

or

t2= (7-sqrt(7^2-4*2*3))/(2*2)

t1= (7+sqrt(25))/(4)

t1= (7+5)/(4)

t1= 3

t2= (7-sqrt(7^2-4*2*3))/(2*2)

t2= (7-sqrt(25))/(4)

t2= (7-5)/(4)

t2= 0.5

But -1<=cosx<=1 ; hence cosx cannot be 3.

Then cosx= 0.5

The primary solution is x= pi/3

The general solution is x=2*n*pi+-(pi/3)

Approved by eNotes Editorial Team `2cos^2 x - 7cos x + 3 = 0`

>> Consider the trigonometric equation as a quadratic expression.

>> Then, factor.

`(2cos x - 1) (cos x - 3) = 0`

>> Set each factor equal to zero.

`(2cos x - 1 ) =0`                 and            `(cos x - 3) =0`

`2cos x = 1`                                               `cos x = 3`

` cos x = 1/2`

>> To solve for the angle x, ignore the factor cos x= 3. Note that the Range of a Cosine function is from -1 to 1 only. Hence, consider only the first factor.

`cos x = 1/2`

>> To determine the value of x, refer to Unit Circle Chart or Table of Trigonometric Function for Special Angles.

`x = 60 deg and 300 deg`

>> Since there is no indicated interval for the angle, then the general solution is:

Answer: `x_1 = 60 + k(360) `  degrees  and

`x_2 = 300 + k(360)`  degrees

Approved by eNotes Editorial Team