Question

solve the equation for solutions over the interval 0°, 360°).
\\(cot 0 + 3 csc 0 = 7\\)
select the correct choice below and, if necessary, fill in the
answer box to complete your choice.
a. the solution set is {}
(type your answer in degrees. do not include the
degree symbol in your answer. round to one decimal
place as needed. use a comma to separate answers
as needed.)
b. the solution is the empty set

Explanation:

Step1: Express in terms of sine and cosine

Recall that \(\cot\theta=\frac{\cos\theta}{\sin\theta}\) and \(\csc\theta = \frac{1}{\sin\theta}\).
The equation \(\cot\theta+3\csc\theta = 7\) becomes \(\frac{\cos\theta}{\sin\theta}+\frac{3}{\sin\theta}=7\).
Since \(\sin\theta
eq0\) (because \(\csc\theta\) is defined), we can multiply through by \(\sin\theta\) to get \(\cos\theta + 3=7\sin\theta\).

Step2: Square both sides

Square the equation \(\cos\theta+3 = 7\sin\theta\).
\((\cos\theta + 3)^2=(7\sin\theta)^2\).
Expand: \(\cos^{2}\theta+6\cos\theta + 9 = 49\sin^{2}\theta\).
Use the identity \(\sin^{2}\theta=1-\cos^{2}\theta\).
\(\cos^{2}\theta+6\cos\theta + 9=49(1 - \cos^{2}\theta)\).
\(\cos^{2}\theta+6\cos\theta + 9=49-49\cos^{2}\theta\).
\(50\cos^{2}\theta+6\cos\theta - 40 = 0\).
Divide by \(2\): \(25\cos^{2}\theta+3\cos\theta - 20=0\).
Let \(x = \cos\theta\), then \(25x^{2}+3x - 20=0\).
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 25\), \(b = 3\), \(c=-20\).
\(x=\frac{-3\pm\sqrt{9+2000}}{50}=\frac{-3\pm\sqrt{2009}}{50}\).
\(\cos\theta=\frac{-3\pm\sqrt{2009}}{50}\).
\(\cos\theta=\frac{-3 + 44.82}{50}\approx0.836\) or \(\cos\theta=\frac{-3-44.82}{50}\approx - 0.956\).

Step3: Find \(\theta\) values

If \(\cos\theta\approx0.836\), then \(\theta=\cos^{-1}(0.836)\approx33.3^{\circ}\) or \(\theta = 360 - 33.3=326.7^{\circ}\).
If \(\cos\theta\approx - 0.956\), then \(\theta=\cos^{-1}(-0.956)\approx163.1^{\circ}\) or \(\theta = 196.9^{\circ}\).
Check these solutions in the original equation \(\cot\theta+3\csc\theta = 7\).
For \(\theta = 33.3^{\circ}\):
\(\cot(33.3)=\frac{\cos(33.3)}{\sin(33.3)}\approx1.5\), \(\csc(33.3)=\frac{1}{\sin(33.3)}\approx1.8\).
\(\cot(33.3)+3\csc(33.3)\approx1.5 + 3\times1.8=1.5+5.4 = 6.9\approx7\).
For \(\theta = 163.1^{\circ}\):
\(\cot(163.1)=\frac{\cos(163.1)}{\sin(163.1)}\approx - 3.3\), \(\csc(163.1)=\frac{1}{\sin(163.1)}\approx3.4\).
\(\cot(163.1)+3\csc(163.1)\approx-3.3+3\times3.4=-3.3 + 10.2=6.9\approx7\).

Answer:

\(33.3,163.1,196.9,326.7\)