To solve the equation:
[tex]-15\csc ^2x-1=-32\cot x[/tex]
We meed to remember the identity:
[tex]\csc ^2x=\cot ^2x+1[/tex]
Plugging this identity in the equation we have:
[tex]\begin{gathered} -15(\cot ^2x+1)-1=-32\cot x \\ -15\cot ^2x-15-1=-32\cot x \\ 15\cot ^2x-32\cot x+16=0 \end{gathered}[/tex]
Hence we have the quadratic equation in the cotangent:
[tex]15\cot ^2x-32\cot x+16=0[/tex]
To solve it let:
[tex]w=\cot x[/tex]
Then we have the quadratic equation:
[tex]15w^2-32w+16=0[/tex]
let's use the general formula to solve it:
[tex]\begin{gathered} w=\frac{-(-32)\pm\sqrt[]{(-32)^2-4(15)(16)}}{2(15)} \\ =\frac{32\pm\sqrt[]{1024-960}}{30} \\ =\frac{32\pm\sqrt[]{64}}{30} \\ =\frac{32\pm8}{30} \\ \text{then} \\ w=\frac{32+8}{30}=\frac{40}{30}=\frac{4}{3} \\ \text{ or } \\ w=\frac{32-8}{30}=\frac{24}{30}=\frac{4}{5} \end{gathered}[/tex]
Once we know the value of w we can find the value of x, remember the definition of w, then we have:
[tex]\begin{gathered} \cot x=\frac{4}{3} \\ \text{ and} \\ \cot x=\frac{4}{5} \end{gathered}[/tex]
Since it is easier to work with the tangent function we will use the fact that:
[tex]\tan x=\frac{1}{\cot x}[/tex]
Hence our equations take the form:
[tex]\begin{gathered} \tan x=\frac{3}{4} \\ \text{and} \\ \tan x=\frac{5}{4} \end{gathered}[/tex]
Finally to solve the equations we need to remember that the tangent function has a period of pi, therefore we have that:
[tex]\begin{gathered} x=\tan ^{-1}(\frac{3}{4})+\pi n \\ \text{and} \\ x=\tan ^{-1}(\frac{5}{4})+\pi n \end{gathered}[/tex]
where n is any integer number. To find the solutions in the interval given we plug n=0 and n=1 in each expression for x; therefore, the solutions in the interval are:
[tex]\begin{gathered} x=0.6435 \\ x=0.8961 \\ x=3.7851 \\ x=4.0376 \end{gathered}[/tex]
