Trigonometric Equation Solutions 4
Find the number of solutions to
\[\frac{1}{\sin^2 \theta} - \frac{1}{\cos^2 \theta} - \frac{1}{\tan^2 \theta} - \frac{1}{\cot^2 \theta} - \frac{1}{\sec^2 \theta} - \frac{1}{\csc^2 \theta} = -3\]in the interval $ 0 \le \theta \le 2 \pi $.
- 1
- 2
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- +
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- -
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- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
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- $[$
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- $\cap$
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- $\infty$
Solution
We can write
\begin{align*}
&\frac{1}{\sin^2 \theta} - \frac{1}{\cos^2 \theta} - \frac{1}{\tan^2 \theta} - \frac{1}{\cot^2 \theta} - \frac{1}{\sec^2 \theta} - \frac{1}{\csc^2 \theta} \\
&= \frac{1}{\sin^2 \theta} - \frac{1}{\cos^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta} - \tan^2 \theta - \cos^2 \theta - \sin^2 \theta \\
&= \frac{1 - \cos^2 \theta}{\sin^2 \theta} - \frac{1}{\cos^2 \theta} - \tan^2 \theta - 1 \\
&= \frac{\sin^2 \theta}{\sin^2 \theta} - \frac{1}{\cos^2 \theta} - \tan^2 \theta - 1 \\
&= -\frac{1}{\cos^2 \theta} - \tan^2 \theta = -3.\end{align*}Hence,
\[\frac{1}{\cos^2 \theta} + \tan^2 \theta = 3.\]We can write this as
\[\frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta} + \tan^2 \theta = 3,\]so $ \tan^2 \theta + 1 + \tan^2 \theta = 3, $ which simplifies to $ \tan^2 \theta = 1 $. Note that up to this point, all our steps have been reversible.
Then $ \tan \theta = \pm 1, $ so the solutions are $ \frac{\pi}{4}, $ $ \frac{3 \pi}{4}, $ $ \frac{5 \pi}{4}, $ and $ \frac{7 \pi}{4}, $ for a total of $ \boxed{4} $ solutions.
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