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Step 1: Rewrite the given equation
The original equation is
\log_{ \frac{1}{2} } \lvert \sin x \rvert = 2 - \log_{ \frac{1}{2} } \lvert \cos x \rvert.
Step 2: Combine the logarithmic terms
Move all the logarithms to one side:
\log_{ \frac{1}{2} } \lvert \sin x \rvert + \log_{ \frac{1}{2} } \lvert \cos x \rvert = 2.
Using properties of logarithms,
\log_{ \frac{1}{2} } \bigl\lvert \sin x \cos x \bigr\rvert = 2.
Step 3: Convert the logarithmic form to exponential form
Recall that \log_{a} (y) = b implies y = a^b. Hence,
\bigl\lvert \sin x \cos x \bigr\rvert = \left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}.
Step 4: Express in terms of \sin 2x
Note that \sin(2x) = 2 \sin x \cos x. Thus,
2 \bigl\lvert \sin x \cos x \bigr\rvert = \bigl\lvert \sin 2x \bigr\rvert = 2 \times \tfrac{1}{4} = \tfrac{1}{2}.
Therefore,
\lvert \sin 2x \rvert = \tfrac{1}{2}.
Step 5: Solve for 2x in the given interval
The equation \sin 2x = \pm \tfrac{1}{2} implies
2x = \sin^{-1}\left(\tfrac{1}{2}\right) \quad \text{or} \quad 2x = \sin^{-1}\left(-\tfrac{1}{2}\right).
Over one full cycle of 2x from 0 to 4\pi (since x \in [0, 2\pi] means 2x \in [0, 4\pi] ),
\sin 2x = \tfrac{1}{2} has solutions:
2x = \frac{\pi}{6}, \; \frac{5\pi}{6}, \; \frac{13\pi}{6}, \; \frac{17\pi}{6},
and
\sin 2x = -\tfrac{1}{2} has solutions:
2x = \frac{7\pi}{6}, \; \frac{11\pi}{6}, \; \frac{19\pi}{6}, \; \frac{23\pi}{6}.
Each corresponds to a distinct x in [0, 2\pi].
Step 6: Determine the total number of solutions
We have a total of 8 distinct values of x in the interval [0, 2\pi].
Final Answer:
The number of distinct solutions in [0, 2\pi] is 8.
