# what FOUR values of theta satisfy the following equation: `7cot^2theta+6cottheta-8=0` on: 0deg <= theta < 360deg express answers in degrees, rounded to two decimal places.

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You should come up with the following substitution, such that:

`cot theta = t`

Replacing t for `cos theta` yields:

`7t^2 + 6t - 8 = 0`

Using quadratic formula yields:

`t_(1,2) = (-6 +- sqrt(36 + 224))/14`

`t_(1,2) = (-6 +- 2sqrt65)/14 =>`

` t_(1,2) = (-3 +- sqrt65)/7 `

`t_1 = 0.723, t_2 = -1.580`

`cot theta = 0.723 =>theta = cot^(-1)0.723 => theta_1 = 0.94 rad`

Since cotangent function is also positive in quadrant 3, yields:

`theta_2 = 3.14 + 0.94 = 4.08 rad`

`cot theta = -1.580 => theta = 3.14 - 0.56 => theta_3 = 2.58 rad`

Since cotangent function is also negative in quadrant 4, yields:

`theta_4 = 2*3.14 - 0.56 = 5.72 rad`

**Hence, evaluating the requested four values yields **`theta_1 = 0.94 rad, theta_2 = 4.08 rad, theta_3 = 2.58 rad, theta_4 = 5.72 rad.`

Solve `7cot^2theta+6cot theta-8=0` on `0<= theta <= 360^@` :

Using the quadratic formula we find `cot theta=(-3+-sqrt(65))/7`

Thus we have `cot theta~~.723` or `cot theta~~-1.58`

(a) `cot theta ~~ .723` ==> `tan theta ~~1.383`

`theta=tan^(-1)1.383~~54.13^@`

The tangent and cotangent are positive in the first and third quadrant so we also have `theta~~234.13^@`

(b) `cot theta~~-1.58` ==> `tan theta ~~-.633`

`theta=tan^(-1)(-.633)~~-32.33^@` . Using this as the reference angle, and knowing that tangent and cotangent are negative in the second and fourth quadrants we get:

`theta~~147.67^@"or"theta~~147.67^@`

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The solutions are `theta~~54.13^@,147.67^@,234.13^@,327.67^@`

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