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$\frac{\log _{10}\left(2\right)}{\log _{10}\left(\sin \left(x\right)\right)}\le \frac{\log _{10}\left(4\sin ^2\left(x\right)\right)}{\log _{10}\left(\sin \left(x\right)\right)}$

From the title. Not homework.

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up vote 1 down vote accepted

On the surface this thing looks like a beast but with a few simple tweaks we can get at the answer. Let's start by looking at the problem, $$\frac{\log _{10}\left(2\right)}{\log _{10}\left(\sin \left(x\right)\right)}\le \frac{\log _{10}\left(4\sin ^2\left(x\right)\right)}{\log _{10}\left(\sin \left(x\right)\right)}$$

We see the denominator is the same so we can cancel them out leaving,

$${\log _{10}\left(2\right)\le \log _{10}\left(4\sin ^2\left(x\right)\right)}$$

Since both sides are logs with base '10' that means the same thing as the left side saying,

$$10^\left(something\right)=2$$ and the right, $$10^\left(something\right)=4\sin^2(x)$$

We can start with assuming for an initial condition where the $10^\left(somethings\right)$ are equal so we are left with, $$2=4\sin^2(x)$$

Now all we have to do is put the inequality back into the mix and solve for 'x', but as Peter Woolfitt pointed out our denominator is negative so our inequality needs flipped, $$2\ge4\sin^2(x)$$ which leads to, $$\frac{1}{2}\ge\sin^2(x)$$ and then get rid of the squared term, $$\frac{1}{\sqrt2}\ge\sin(x)$$

Now for this last part we want to solve for 'x' but as pointed out by JChau we have to be careful.

We are looking for values for 'x' (or $\theta$) where the value of $sin(\theta)$ is less than or equal to $\frac{1}{\sqrt2}$. Thinking about the unit circle and where $\sin\frac{\Pi}{4}$ lies (and that $\sin\theta$ is a representation of the 'y' values on the unit circle) we need our equality to hold true for all $\theta$ where the 'y' value is less than $\frac{1}{\sqrt2}$ which would be all $\theta$ sweeping clockwise from $\frac{3\Pi}{4}$ to $\frac{\Pi}{4}$

So now we need to write this out, $$\frac{\Pi}{4}\ge x\ge\frac{3\Pi}{4}$$

But this still doesn't quite solve the problem because on a unit circle we can make angles that can go up to infinity and each time $\sin(\theta)$ is in this same 'range' the answer will be true. So for this we add in the following, $$\frac{\Pi}{4}+2\Pi[k] \ge x\ge\frac{3\Pi}{4}+2\Pi[k]$$

The $+2\Pi$ takes care of the angle, and the [k] represents any Real integer for 'k' so that we can cover all angles up to $\sin(\theta*\inf)$ and $\sin(-\theta*\inf)$. I should add there is a 'correct' way to write this and this is not it.

As you can see it doesn't take much to simplify the problem initially to terms we can more easily solve for (just have to be careful about where the inequality is true and which direction it points, it's the simple things that get you:)

Hope this helps!

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You need to reverse the direction of inequality when canceling since you are canceling something negative – Peter Woolfitt Aug 13 '14 at 23:09
You guys are awesome, thanks for the tips! – eatscrayons Aug 13 '14 at 23:45

$$\frac{\log _{10}\left(2\right)}{\log _{10}\left(\sin \left(x\right)\right)}\le \frac{\log _{10}\left(4\sin ^2\left(x\right)\right)}{\log _{10}\left(\sin \left(x\right)\right)} $$

So that the logarithm is well defined it should be:

$$\sin \left(x\right)>0 \Rightarrow 2 \pi n <x<2 \pi n +\pi, n \in \mathbb{Z} \ \ \ (1)$$

Since $0< \sin \left(x\right) \leq 1$, $\log _{10}\left(\sin \left(x\right)\right)<0$.

Therefore, multiplying both sides of the inequality by $\log _{10}\left(\sin \left(x\right)\right)$ we get the following:

$$\log _{10}\left(2\right) \geq \log _{10}\left(4\sin ^2\left(x\right)\right) \\ \Rightarrow 10^{\log _{10}\left(2\right)} \geq 10^{\log _{10}\left(4\sin ^2\left(x\right)\right)} \\ \Rightarrow 2 \geq 4 \sin^2\left(x\right) \\ \Rightarrow \frac{1}{2} \geq \sin^2\left(x\right) \\ \Rightarrow \frac{1}{\sqrt{2}} \geq \left | \sin\left(x\right) \right | \\ \Rightarrow \sin\left(x\right) \geq -\frac{1}{\sqrt{2}} \text{ or } \sin\left(x\right) \leq \frac{1}{\sqrt{2}}$$

The first inequality stands since it should be $\sin\left(x\right)>0$.

Therefore, we have to find for which $x$ the second inequality stands.

$$\sin\left(x\right) \leq \frac{1}{\sqrt{2}} \Rightarrow 2 \pi n-\frac{5}{4} \pi \leq x \leq 2 \pi n +\frac{\pi}{4}, n \in \mathbb{Z} \\ \Rightarrow 2 \pi (n-1)+\frac{3}{4} \pi \leq x \leq 2 \pi n +\frac{\pi}{4}, n \in \mathbb{Z} \ \ \ (2)$$

So, to solve the original inequality you have to find which $x$ satisfy simultaneously the relations $(1)$ and $(2)$.

$$2 \pi n <x \leq 2 \pi n +\frac{\pi}{4} \text{ and } 2 \pi n+\frac{3}{4} \pi \leq x < 2 \pi n+\pi, n \in \mathbb{Z}$$

Therefore, the inequality stands for $\displaystyle{x \in (2 \pi n,2 \pi n +\frac{\pi}{4}] \cup [ 2 \pi n+\frac{3}{4} \pi,2 \pi n+\pi), n \in \mathbb{Z}}$

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We first assume $x\in(2\pi n,2\pi n+\pi)$ and $x\ne\dfrac{\pi}{2}+2\pi n$ so that $\dfrac{1}{\log(\sin x)}$ makes sense. Now, since $0<\sin x<1$, we have that $\log(\sin x)<0$. Hence multiplying both sides by $\log(\sin x)$ reverses the inequality and the rest is easy.

$$\begin{align} &\frac{\log2}{\log(\sin x)}\le\frac{\log(4\sin^2x)}{\log(\sin x)} \\\iff&\log2\ge\log(4\sin^2x) \\\iff&2\ge4\sin^2x \\\iff&\frac{1}{\sqrt{2}}\ge\sin x \end{align}$$

Note that in the last line we are implicitly using $0<\sin x$.

Hence $x\in(2\pi n,2\pi n+\frac{\pi}{4}]\cup[2\pi n+\frac{3\pi}{4},2\pi n+\pi)$.

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Thanks! This drove me crazy :( – Tomas Smith Aug 13 '14 at 23:19
@TomasSmith Glad to help – Peter Woolfitt Aug 13 '14 at 23:21

$$\frac{\log(2)}{\log(\sin x)}\leq \frac{4\sin^2 x}{\log(\sin x)}$$ Multiply both sides by $\log(\sin x)$, reversing the inequality and getting rid of denominators. $$\log(2)\geq \log(4\sin^2 x)$$ Get rid of the $\log$s. $$2\geq 4\sin^2 x$$ $$\frac 12\geq \sin^2 x$$ $$\frac 1{\sqrt 2}\geq \sin x$$ Here is a graph of the inequality:

enter image description here

The red line is the graph of $y=\sin x$, and the blue line is the graph of $y=\frac 1{\sqrt 2}$.

The solutions are the regions where the red line is below the blue line (including when it touches the blue line). There are an infinite amount of solutions, but there is a general solution.

The general solution is:

$$x\in \left(2\pi n, 2\pi n+\frac{\pi}4\right]\cup \left[2\pi n+\frac{3\pi}{4}, 2\pi n+\pi\right), \ n\in \mathbb Z$$

If you are unfamiliar with interval notation, then basically the solutions for $x$ are:

$$2\pi n< x\leq 2\pi n +\frac{\pi}4 \ \text{AND} \ 2\pi n +\frac {3\pi}4\leq x<2\pi n+\pi, \ n\in \mathbb Z$$

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Notice that the inequality can only make sense for $x$ so that $0<\sin(x)<1$. In this case the inequality seems equivalent to


by the change of base formula. So:


since $0<\sin(x)<1$ (which means the function $y\to\sin(x)^{y}$ is decreasing) and hence: $$\frac{1}{\sqrt{2}}\ge\lvert \sin(x)\rvert=\sin(x)$$

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