1. ## Trigonometric Inequality Problem

Hi, this trigonometric inequality was on our test, and I didn't know how to solve it.

Solve:

sin(pi/2(x)) < (x-2)^3-1

The cycle for hte sine function is 1, and I know one of the points of intersection is 3, but I do not know how to solve for the other two points of intersection for the inequality.

2. Originally Posted by KelvinScale
Hi, this trigonometric inequality was on our test, and I didn't know how to solve it.

Solve:

sin(pi/2(x)) < (x-2)^3-1

The cycle for hte sine function is 1, and I know one of the points of intersection is 3, but I do not know how to solve for the other two points of intersection for the inequality.

is this correct?

$\displaystyle \sin\left(\frac{\pi}{2x}\right) < (x-2)^3-1$

3. No it's (pi/2)(x) with the x on top, not the bottom.

4. Originally Posted by KelvinScale
No it's (pi/2)(x) with the x on top, not the bottom.
like this..

$\displaystyle \sin\left(\frac{\pi x}{2}\right) < (x-2)^3-1$

5. yes, I think you can solve it by guess and check, but I was wondering if there was a systematic way of doing it.

6. my assumption on this. is that since it is an inequality you have 2 graphs
and the answer is a particular area between the two

you might want to look at this.....
make sure I have plugged it in correct

http://www.wolframalpha.com/input/?i=+sin((x*pi)%2F2)+%3C+(x-2)^3-1

hope this helps...

7. Originally Posted by bigwave
my assumption on this. is that since it is an inequality you have 2 graphs
and the answer is a particular area between the two

you might want to look at this.....
make sure I have plugged it in correct

http://www.wolframalpha.com/input/?i=+sin((x*pi)%2F2)+%3C+(x-2)^3-1

hope this helps...
the answer has nothing to do with the area between the two. From the graph we can see that for (approximately) $\displaystyle x > 2.587$, the cubic exceeds the sine function, and hence is greater on that interval. It looks like it would be pretty annoying to find the exact value for the left end-point of the interval (if there even is one!), to find this value, you can use some kind of approximation method, like the Newton-Raphson method--if you're a glutton for punishment.