1. ## Inequalities

I was asked to solve $\frac{x^2-x-2}{x-1}<2$ from which the solution is $x<0$ or $1

The next part asks: Hence solve $\frac{cos^22\theta-cos2\theta-2}{cos2\theta-1}<2$, where $0 &\leq \theta &\leq 2\pi$

I tried doing: Replace $x$ with $cos2\theta$

$cos2\theta<0$ or $1

therefore, $\theta<\frac{\pi}{4}$ or $0<\theta<$ but $cos^{-1}3=$error

Can you show me the proper way to solve such an inequality? I totally forgot how to tackle this...

2. ## Re: Inequalities

If $0\leq\theta\leq 2\pi$, then $\cos 2\theta<0$ is equivalent to $\pi/4<\theta<3\pi/4$. Draw the unit circle and see for which angles cosine is negative.

Concerning $1<\cos2\theta<3$, it helps to draw a graph of $\cos2\theta$ and see which portion is located between the lines y = 1 and y = 3. Can $\cos2\theta$ ever be greater than 1?

3. ## Re: Inequalities

Originally Posted by emakarov
If $0\leq\theta\leq 2\pi$, then $\cos 2\theta<0$ is equivalent to $\pi/4<\theta<3\pi/4$. Draw the unit circle and see for which angles cosine is negative.

Concerning $1<\cos2\theta<3$, it helps to draw a graph of $\cos2\theta$ and see which portion is located between the lines y = 1 and y = 3. Can $\cos2\theta$ ever be greater than 1?
no

4. ## Re: Inequalities

That's right. I hope you have the answer to the original in equality.

5. ## Re: Inequalities

I drew the circle and it shows that the 2nd and 3rd quadrant are where $cos\theta<0$ but the question is $cos2\theta$, shouldn't there be some changes? why is it still the same?

6. ## Re: Inequalities

It is not the same. As you said, we have $\cos\alpha<0$ iff $\pi/2<\alpha<3\pi/2$. In particular, for $\alpha=2\theta$, $\cos2\theta<0$ iff $\pi/2<2\theta<3\pi/2$ iff $\pi/4<\theta<3\pi/4$.

7. ## Re: Inequalities

nothing lies between the line y=1 and y=3, do i reject this solution? but it seems that the model answer has another solution of $\frac{5\pi}{4}<\theta<\frac{7\pi}{4}$ in addition to the one you mentioned

8. ## Re: Inequalities

Originally Posted by Punch
nothing lies between the line y=1 and y=3, do i reject this solution?
Yes, no value of $\theta$ can contribute to the solution of $1<\cos2\theta$.

Originally Posted by Punch
but it seems that the model answer has another solution of $\frac{5\pi}{4}<\theta<\frac{7\pi}{4}$ in addition to the one you mentioned
You are right. I missed that $5\pi/2<2\theta<7\pi/2$ also implies $\cos2\theta<0$, and even though the interval $(5\pi/2,7\pi/2)$ is outside of $[0,2\pi]$, the corresponding interval for $\theta$: $(5\pi/4,7\pi/4)$ is inside $[0,2\pi]$.

It is better to be precise and to select the solutions from $[0,2\pi]$ in the end. So, $\cos2\theta$ iff $\pi/2+2\pi k<2\theta<3\pi/2+2\pi k$, $k\in\mathbb{Z}$. This is equivalent to $\pi/4+\pi k<\theta<3\pi/4+\pi k$, $k\in\mathbb{Z}$. Of these, only intervals for k = 0 and k = 1 lie inside $[0,2\pi]$.