Question

Use trigonometric identities, algebraic methods, and inverse trigonometric functions, as necessary, to solve the following trigonometric equation on the interval [0, 28t).Round your answer to four decimal places, if necessary. If there is no solution, indicate "No Solution."- 15csc?(x) - 1 = -32cot(x)yea

Answer 1

To solve the equation:

$-15\csc ^2x-1=-32\cot x$

We meed to remember the identity:

$\csc ^2x=\cot ^2x+1$

Plugging this identity in the equation we have:

$\begin{gathered} -15(\cot ^2x+1)-1=-32\cot x \\ -15\cot ^2x-15-1=-32\cot x \\ 15\cot ^2x-32\cot x+16=0 \end{gathered}$

Hence we have the quadratic equation in the cotangent:

$15\cot ^2x-32\cot x+16=0$

To solve it let:

$w=\cot x$

Then we have the quadratic equation:

$15w^2-32w+16=0$

let's use the general formula to solve it:

$\begin{gathered} w=\frac{-(-32)\pm\sqrt[]{(-32)^2-4(15)(16)}}{2(15)} \\ =\frac{32\pm\sqrt[]{1024-960}}{30} \\ =\frac{32\pm\sqrt[]{64}}{30} \\ =\frac{32\pm8}{30} \\ \text{then} \\ w=\frac{32+8}{30}=\frac{40}{30}=\frac{4}{3} \\ \text{ or } \\ w=\frac{32-8}{30}=\frac{24}{30}=\frac{4}{5} \end{gathered}$

Once we know the value of w we can find the value of x, remember the definition of w, then we have:

$\begin{gathered} \cot x=\frac{4}{3} \\ \text{ and} \\ \cot x=\frac{4}{5} \end{gathered}$

Since it is easier to work with the tangent function we will use the fact that:

$\tan x=\frac{1}{\cot x}$

Hence our equations take the form:

$\begin{gathered} \tan x=\frac{3}{4} \\ \text{and} \\ \tan x=\frac{5}{4} \end{gathered}$

Finally to solve the equations we need to remember that the tangent function has a period of pi, therefore we have that:

$\begin{gathered} x=\tan ^{-1}(\frac{3}{4})+\pi n \\ \text{and} \\ x=\tan ^{-1}(\frac{5}{4})+\pi n \end{gathered}$

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:

$\begin{gathered} x=0.6435 \\ x=0.8961 \\ x=3.7851 \\ x=4.0376 \end{gathered}$