how many solutions of the equation x^4x^3*sinxx^2*cosx=0?

 Posts: 9
 Joined: Sat Oct 01, 2011 11:21 pm
how many solutions of the equation x^4x^3*sinxx^2*cosx=0?
how to solve this kind of problem? tks in advance
Re: how many solutions of the equation x^4x^3*sinxx^2*cosx=0?
There are 3 unique solutions (without counting multiplicities).
Notice that the lefthand side factors as x^2 * (x^2  x sinx  cosx). The x^2 factor gives you a solution for x = 0. Let's focus on the other factor, f(x) = x^2  x sinx  cosx. Notice that f is negative around a neighborhood of the origin and approaches + infinity as x gets large (since the x^2 term dominates). Hence f has at least two roots as its sign changes. Does it have any more?
Let's look at its critical points. f'(x) = 2x  x cosx = x * (2  cosx) = 0 only at x = 0. (That's a local minimum of f because f' < 0 for x < 0 and f' > 0 for x > 0.) Since f has only a single critical point, it will have at most two roots (there's at least one critical point between any two roots). But we already found two roots. So we are done.
Notice that the lefthand side factors as x^2 * (x^2  x sinx  cosx). The x^2 factor gives you a solution for x = 0. Let's focus on the other factor, f(x) = x^2  x sinx  cosx. Notice that f is negative around a neighborhood of the origin and approaches + infinity as x gets large (since the x^2 term dominates). Hence f has at least two roots as its sign changes. Does it have any more?
Let's look at its critical points. f'(x) = 2x  x cosx = x * (2  cosx) = 0 only at x = 0. (That's a local minimum of f because f' < 0 for x < 0 and f' > 0 for x > 0.) Since f has only a single critical point, it will have at most two roots (there's at least one critical point between any two roots). But we already found two roots. So we are done.

 Posts: 9
 Joined: Sat Oct 01, 2011 11:21 pm
Re: how many solutions of the equation x^4x^3*sinxx^2*cosx=0?
thank you very much!! now I think I conquer this kind of problems!
I can find the solution of another similar question by myself (cheers!!)
Quesion is y=logx and y=cx^4 have only one intersection, what is the value of c?
The answer is 1/4e.
let f(x)=logxcx^4=0 which means f(x) has only one intersction with xaxis.
f'(x)=1/x4cx^3=0
x^4=square1/4c
let c>0
x>square1/4c, f'(x)<0 f(x) go decrease
x<square1/4c, f'(x)>0, f(x) go increase
so f(x=square1/4c) must be 0 so that f(x) has only one intersection with xaxis.
I can find the solution of another similar question by myself (cheers!!)
Quesion is y=logx and y=cx^4 have only one intersection, what is the value of c?
The answer is 1/4e.
let f(x)=logxcx^4=0 which means f(x) has only one intersction with xaxis.
f'(x)=1/x4cx^3=0
x^4=square1/4c
let c>0
x>square1/4c, f'(x)<0 f(x) go decrease
x<square1/4c, f'(x)>0, f(x) go increase
so f(x=square1/4c) must be 0 so that f(x) has only one intersection with xaxis.