Solve in the interval 0^@, 360^@ 2csc^2x + cotx = 10 How?

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Oct 24, 2017

${\cot}^{- 1} \left(\frac{1 + \sqrt{65}}{4}\right) \approx {23.816}^{\circ}$

${\cot}^{- 1} \left(\frac{1 - \sqrt{65}}{4}\right) \approx {150.473}^{\circ}$

${\cot}^{- 1} \left(\frac{1 + \sqrt{65}}{4}\right) + {180}^{\circ} \approx {203.816}^{\circ}$

${\cot}^{- 1} \left(\frac{1 - \sqrt{65}}{4}\right) + {180}^{\circ} \approx {330.473}^{\circ}$

Explanation:

Note that:

${\csc}^{2} x = \frac{1}{\sin} ^ 2 x = \frac{{\cos}^{2} x + {\sin}^{2} x}{\sin} ^ 2 x = 1 + {\cot}^{2} x$

So:

$10 = 2 {\csc}^{2} x + \cot x$

$\textcolor{w h i t e}{10} = 2 \left({\cot}^{2} x + 1\right) + \cot x$

$\textcolor{w h i t e}{10} = 2 {\cot}^{2} x + \cot x + 2$

Subtract $10$ from both ends to get:

$2 {\cot}^{2} x + \cot x - 8 = 0$

This is in the form:

$a {t}^{2} + b t + c = 0$

with $t = \cot x$, $a = 2$, $b = 1$ and $c = - 8$.

So using the quadratic formula, we have:

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

$\textcolor{w h i t e}{\cot x} = \frac{- \textcolor{b l u e}{1} \pm \sqrt{{\textcolor{b l u e}{1}}^{2} - 4 \left(\textcolor{b l u e}{2}\right) \left(\textcolor{b l u e}{- 8}\right)}}{2 \left(\textcolor{b l u e}{2}\right)}$

$\textcolor{w h i t e}{\cot x} = \frac{- 1 \pm \sqrt{65}}{4}$

Note that (expressed in degrees):

• $\cot x$ is continuous and monotonically decreasing on $\left({0}^{\circ} , {180}^{\circ}\right)$ with range $\left(- \infty , \infty\right)$.

• The range of ${\cot}^{- 1} \left(y\right)$ is $\left({0}^{\circ} , {180}^{\circ}\right)$.

• $\cot x$ has period ${180}^{\circ}$.

Hence all the solutions in $\left({0}^{\circ} , {360}^{\circ}\right)$ are:

${\cot}^{- 1} \left(\frac{1 + \sqrt{65}}{4}\right) \approx {23.816}^{\circ}$

${\cot}^{- 1} \left(\frac{1 - \sqrt{65}}{4}\right) \approx {150.473}^{\circ}$

${\cot}^{- 1} \left(\frac{1 + \sqrt{65}}{4}\right) + {180}^{\circ} \approx {203.816}^{\circ}$

${\cot}^{- 1} \left(\frac{1 - \sqrt{65}}{4}\right) + {180}^{\circ} \approx {330.473}^{\circ}$

Note that your calculator may give a negative value for ${\cot}^{- 1} \left(\frac{1 - \sqrt{65}}{4}\right)$ (namely $\approx - {29.527}^{\circ}$). If it does then just add ${180}^{\circ}$.

If your calculator does not have ${\cot}^{- 1}$ then take the reciprocal and use ${\tan}^{- 1}$, but note that this will give values in the range $\left(- {90}^{\circ} , {90}^{\circ}\right)$, so you will need to add an initial ${180}^{\circ}$ to the $\approx - {29.527}^{\circ}$ value to get a solution in the requested range.

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