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.
Your help is greatly appreciated!
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) , 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.