# How do you use the quadratic formula to solve 3csc^2theta-2csctheta=2 for 0<=theta<360?

Oct 19, 2017

see below

#### Explanation:

let $u = \csc \theta$ then $3 {\csc}^{2} \theta - 2 \csc \theta = 2$ becomes $3 {u}^{2} - 2 u = 2$

then if we put everything on one side we have $3 {u}^{2} - 2 u - 2 = 0$. So let's use the quadratic formula to solve.

Now take $a = 3 , b = - 2 , \mathmr{and} c = - 2$ and put it in to the

formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$u = \frac{2 \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(3\right) \left(- 2\right)}}{2 \left(3\right)}$

$= \frac{2 \pm \sqrt{28}}{6}$

$= \frac{2 \pm 2 \sqrt{7}}{6}$

$= \frac{\cancel{2} \left(1 \pm \sqrt{7}\right)}{\cancel{6} 3}$

$\csc \theta = \frac{1 \pm \sqrt{7}}{3}$

$\theta = {\csc}^{-} 1 \left(\frac{1 \pm \sqrt{7}}{3}\right) = {\sin}^{-} 1 \left(\frac{3}{1 \pm \sqrt{7}}\right)$

theta=sin^-1 (3/(1+ sqrt(7))) or theta = cancel(sin^-1 (3/(1- sqrt(7)))=undef

$\theta = 55.37370265 + 360 {n}^{\circ} , \mathmr{and} \theta = {180}^{\circ} - 55.37370265 + 360 {n}^{\circ}$

$\theta = {55.37370265}^{\circ} + 360 {n}^{\circ} , \mathmr{and} \theta = {124.6262974}^{\circ} + 360 {n}^{\circ}$

Now let's put in n values (integers) to determine our final answer

$n = 0 , \theta = {55.37}^{\circ} , {124.636}^{\circ}$

Note that if we put in other n-values our answers will be outside of the domain so our solution set is

$S = \left\{{55.37}^{\circ} , {124.636}^{\circ}\right\}$