Question

solve the equation for solutions over the interval 0°, 360°). cotθ + 5cscθ = 6 select the correct choice below and, if necessary, fill in the answer box to complete your choice. oa. the solution set is { (type your answer in degrees. do not include the degree symbol in your answer. round to one decimal pl as needed. use a comma to separate answers as needed.) ob. the solution is the empty set.

Explanation:

Step1: Use trigonometric identities

Recall that \(\cot\theta=\frac{\cos\theta}{\sin\theta}\) and \(\csc\theta = \frac{1}{\sin\theta}\).
The equation \(\cot\theta + 5\csc\theta=6\) becomes \(\frac{\cos\theta}{\sin\theta}+\frac{5}{\sin\theta}=6\).
Since \(\sin\theta
eq0\) (because \(\csc\theta\) is defined), we can multiply through by \(\sin\theta\) to get \(\cos\theta + 5=6\sin\theta\).

Step2: Square both sides

Square both sides of the equation \(\cos\theta+5 = 6\sin\theta\).
\((\cos\theta + 5)^2=(6\sin\theta)^2\).
Expand: \(\cos^{2}\theta + 10\cos\theta+25 = 36\sin^{2}\theta\).
Use the identity \(\sin^{2}\theta=1 - \cos^{2}\theta\).
\(\cos^{2}\theta+10\cos\theta + 25=36(1-\cos^{2}\theta)\).
Expand further: \(\cos^{2}\theta+10\cos\theta + 25=36 - 36\cos^{2}\theta\).
Bring all terms to one side: \(37\cos^{2}\theta+10\cos\theta - 11 = 0\).
Let \(x = \cos\theta\), then the quadratic equation is \(37x^{2}+10x - 11=0\).
Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 37\), \(b = 10\), \(c=-11\).
\(x=\frac{-10\pm\sqrt{100+1628}}{2\times37}=\frac{-10\pm\sqrt{1728}}{74}=\frac{-10\pm12\sqrt{12}}{74}=\frac{-10\pm34.0}{74}\).
We get \(x=\frac{-10 + 34}{74}=\frac{24}{74}=\frac{12}{37}\approx0.324\) or \(x=\frac{-10-34}{74}=\frac{-44}{74}\approx - 0.595\).

Step3: Solve for \(\theta\)

If \(\cos\theta=\frac{12}{37}\), then \(\theta=\cos^{-1}(\frac{12}{37})\approx67.4^{\circ}\) or \(\theta = 360^{\circ}-\cos^{-1}(\frac{12}{37})\approx292.6^{\circ}\).
If \(\cos\theta=-0.595\), then \(\theta=\cos^{-1}(-0.595)\approx126.6^{\circ}\).
Check these solutions in the original equation.
For \(\theta = 67.4^{\circ}\):
\(\cot(67.4^{\circ})+5\csc(67.4^{\circ})=\frac{\cos(67.4^{\circ})}{\sin(67.4^{\circ})}+5\frac{1}{\sin(67.4^{\circ})}\)
\(\sin(67.4^{\circ})\approx0.924\), \(\cos(67.4^{\circ})\approx0.383\)
\(\frac{0.383}{0.924}+5\times\frac{1}{0.924}=\frac{0.383 + 5}{0.924}=\frac{5.383}{0.924}\approx5.83
eq6\) (reject).
For \(\theta = 126.6^{\circ}\):
\(\sin(126.6^{\circ})\approx0.8\), \(\cos(126.6^{\circ})\approx - 0.6\)
\(\frac{-0.6}{0.8}+5\times\frac{1}{0.8}=\frac{-0.6 + 5}{0.8}=\frac{4.4}{0.8}=5.5
eq6\) (reject).
For \(\theta = 292.6^{\circ}\):
\(\sin(292.6^{\circ})\approx - 0.924\), \(\cos(292.6^{\circ})\approx0.383\)
\(\frac{0.383}{-0.924}+5\times\frac{1}{-0.924}=\frac{0.383+5}{-0.924}=\frac{5.383}{-0.924}
eq6\) (reject).

Let's go back to \(\cos\theta+5 = 6\sin\theta\), rewrite it as \(\cos\theta-6\sin\theta=-5\).
The left - hand side can be written as \(R\cos(\theta+\alpha)\) where \(R=\sqrt{1 + 36}=\sqrt{37}\approx6.08\), \(\tan\alpha = 6\).
\(\cos(\theta+\alpha)=-\frac{5}{\sqrt{37}}\approx - 0.825\).
\(\theta+\alpha=\cos^{-1}(-0.825)\approx145.6^{\circ}+360^{\circ}n\) or \(\theta+\alpha = 214.4^{\circ}+360^{\circ}n\)
\(\alpha=\tan^{-1}(6)\approx80.5^{\circ}\)
\(\theta\approx145.6^{\circ}-80.5^{\circ}=65.1^{\circ}\) or \(\theta\approx214.4^{\circ}-80.5^{\circ}=133.9^{\circ}\) (checking \(65.1^{\circ}\)):
\(\cot(65.1^{\circ})+5\csc(65.1^{\circ})=\frac{\cos(65.1^{\circ})}{\sin(65.1^{\circ})}+5\frac{1}{\sin(65.1^{\circ})}\)
\(\sin(65.1^{\circ})\approx0.907\), \(\cos(65.1^{\circ})\approx0.422\)
\(\frac{0.422}{0.907}+5\times\frac{1}{0.907}=\frac{0.422 + 5}{0.907}=\frac{5.422}{0.907}\approx6\)
\(\theta = 65.1^{\circ}\)

Answer:

A. The solution set is \(65.1\)