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

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`2 cos^2 x - 7 cos x + 3 =0`

Ok consider the trigonometric equation as a quadratic expression.

Then,factor.

`(2cosx-1) (cosx-3)=0`

Ok then set each factor equal to zero

`(2cosx-1)=0` and `(cosx-3)=0`

`2cosx=1` `cosx=3`

`cosx =1/2`

To solve for the angle x, ignore the factor cos x=3.Please note that the Range of Cosine function is from -1 to 1 only.

Hence consider only the first factor.

`cosx=1/2`

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

x=60 deg and 300 deg

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

`x_1 =60+k(360) ` degress

`x_2 =300+k(360) ` degress

You should convert the given trigonometric equation into a quadratic equation, using the substitution `cos x = y` , such that:

`2y^2 - 7y + 3 = 0`

Using quadratic formula, yields:

`y_(1,2) = (7+-sqrt(49 - 24))/4`

`y_(1,2) = (7+-sqrt25)/4 => y_(1,2) = (7+-5)/4 => y_1 = 3; y_2 = 1/2`

You need to replace back cos x for y such that:

`cos x = 3` invalid since the values of cosine function cannot be larger than 1

`cos x = 1/2`

Since the cosine values are positive in quadrants 1 and 4 yields:

`x = pi/3` (quadrant 1)

`x = 2pi - pi/3 => x = (5pi)/3` (quadrant 4)

**Hence, evaluating the exactly special angles that make the equation to hold, yields **`x = pi/3, x = (5pi)/3.`