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?

Question

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!

Answer 1

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 $x = \frac{\pi}{4}-2\pi$.

Answer 2

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.