Logarithmic Equation Solutions
If $ 0 \le \theta \le 4 \pi, $ find all values of $ \theta $ which satisfy
\[\log_{\frac{1}{2} \sin 2 \theta} \sin \theta = \frac{1}{2}.\]Enter all the solutions, separated by commas.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
From the given equation,
\[\left( \frac{1}{2} \sin 2 \theta \right)^{\frac{1}{2}} = \sin \theta.\]Squaring both sides, we get
\[\frac{1}{2} \sin 2 \theta = \sin^2 \theta.\]Then $ \sin \theta \cos \theta = \sin^2 \theta, $ so
\[\sin \theta \cos \theta - \sin^2 \theta = \sin \theta (\sin \theta - \cos \theta) = 0.\]Thus, $ \sin \theta = 0 $ or $ \sin \theta = \cos \theta $.
If $ \sin \theta = 0, $ then $ \frac{1}{2} \sin 2 \theta = 0, $ which is not allowed as the base of a logarithm.
Otherwise, $ \sin \theta = \cos \theta $. Then $ \tan \theta = 1 $. The solutions to this equation are $ \frac{\pi}{4}, $ $ \frac{5 \pi}{4}, $ $ \frac{9 \pi}{4}, $ and $ \frac{13 \pi}{4} $. But $ \sin \theta $ must be positive, in order to take the logarithm of it, so the only solutions are $ \boxed{\frac{\pi}{4}, \frac{9 \pi}{4}} $.
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