# Continuity (intermediate value theorem)

mtayab1994

## Homework Statement

Let n be a natural number.
Prove that the equation: $$x^{2}(cos(x))^{n}+xsin(x)+1=0$$ has an infinite amount of solutions.

## The Attempt at a Solution

I named that equation f(x)=0 and I said that f(a)<f(0)<f(b) and that f(a) x f(b) < 0.

Should I choose n=1 and chose a=0 and b=pi ? I know that the equation is a period equation so it will have an infinite amount of solutions, but i'm not sure how to show it because I've only been dealing with equations that want you to show there is at least one solution in [a,b]. Any help would be really appreciated. Thank you.

Homework Helper
Gold Member
Hint: consider the values of $x$ for which $\cos(x) = 0$. What does $\sin(x)$ equal for these values of $x$?

mtayab1994
Hint: consider the values of $x$ for which $\cos(x) = 0$. What does $\sin(x)$ equal for these values of $x$?

When cos(x)=0 then sin(x)=1 for values in the form of x= pi/2+2kpi

So then we could have : f(3pi/2) < f(0) < f(pi/2) ??

Homework Helper
Gold Member
When cos(x)=0 then sin(x)=1 for values in the form of x= pi/2+2kpi
OK, and what about $x = 3\pi/2 + 2k\pi$?

So then we could have : f(3pi/2) < f(0) < f(pi/2) ??
Are you asking me, or telling me? What does f(x) reduce to when $x = \pi/2 + 2k\pi$? How about when $x = 3\pi/2 + 2k\pi$?

mtayab1994
OK, and what about $x = 3\pi/2 + 2k\pi$?

Are you asking me, or telling me? What does f(x) reduce to when $x = \pi/2 + 2k\pi$? How about when $x = 3\pi/2 + 2k\pi$?

Don't mind me I've had a really long day i wanted to say x=pi/2+kpi.

mtayab1994
Ok this is what i got: f(pi/2)=pi/2+1 and f(3pi/2)=1-3pi/2 so we can say that:

f(3pi/2) < 0 < f(pi/2) am i correct?

Homework Helper
Gold Member
Ok this is what i got: f(pi/2)=pi/2+1 and f(3pi/2)=1-3pi/2 so we can say that:

f(3pi/2) < 0 < f(pi/2) am i correct?

Yes, that's right. Now you can apply the intermediate value theorem.

You can argue similarly for $f(3\pi/2 + 2k\pi)$ and $f(\pi/2 + 2k\pi)$, for any positive integer $k$.

mtayab1994
Yes, that's right. Now you can apply the intermediate value theorem.

You can argue similarly for $f(3\pi/2 + 2k\pi)$ and $f(\pi/2 + 2k\pi)$, for any positive integer $k$.

Alright I've got it thanks for your help.