If $\log_{0.5}\sin x=1-\log_{0.5}\cos x$ , then the number of solutions in the interval [$-2\pi, 2\pi$] is? In the given solution,the answer gives only two solutions. However, when a graph of sin 2x is plotted, we see that it attains a value of 1 at four points in the given interval. What am I missing? Thank you!
 A: The book is giving one solution $[-2\pi,2\pi]$ but giving another $(0, \pi/2)$.  It looks like the book is inconsistent, but you are correct in your answer.
The general solution for the equation above is $$\dfrac {\pi}{4} + 2 \pi k, k \in (0, 1)$$.
EDIT: Removed the $\pm$ from the previous solution as this would result in negative logarithms; expanded the domain.
A: Recall that in $\Bbb R$ the property $\log \alpha + \log\beta = \log\alpha\beta$, is only valid when both $\alpha$ and $\beta$ are strictly positive.
The equation
$$\log_{0.5}\sin x=1-\log_{0.5}\cos x$$
is thus equivalent to
$$
\begin{cases}
\sin x> 0\\
\cos x > 0\\
\log_{0.5}\sin x\cos x = 1,
\end{cases}
$$
meaning
$$
\begin{cases}
2k\pi < x < \frac{\pi}{2}+2k\pi\\
\sin x \cos x = \frac{1}{2}
\end{cases}\ ,\ k\in \Bbb Z
$$
or, equivalently,
$$
\begin{cases}
2k\pi < x < \frac{\pi}{2}+2k\pi\\
\sin 2x  = 1.
\end{cases}
$$
Solving with respect to $x$ the second equation yields
$$
\begin{cases}
2k\pi < x < \frac{\pi}{2}+2k\pi\\
x = \frac{\pi}{4}+k\pi
\end{cases}
$$
The required solutions in $[-2\pi, 2\pi]$ therefore are $x = \frac{\pi}{4}$ and $x = \frac{\pi}{4}-2\pi$. 
