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} \\ =(32\pm8)/(30) \\ \text{then} \\ w=(32+8)/(30)=(40)/(30)=(4)/(3) \\ \text{ or } \\ w=(32-8)/(30)=(24)/(30)=(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=(4)/(3) \\ \text{ and} \\ \cot x=(4)/(5) \end{gathered}`

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

`\tan x=(1)/(\cot x)`

Hence our equations take the form:

`\begin{gathered} \tan x=(3)/(4) \\ \text{and} \\ \tan x=(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)((3)/(4))+\pi n \\ \text{and} \\ x=\tan ^(-1)((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}`