Question

Help me pleaze :
Resolve the equation in IR:
cos^5(x)-sin^5(x)=1

Answer 1

Notice that the
\[
2 k \pi \text { for } k=0, \pm1, \pm2, \pm 3, \cdots\\
and
\\
-\frac \pi 2 + 2 k \pi \text { for } k=0, \pm1, \pm2, \pm 3, \cdots\\
\]
are solutions.

Answer 2

How do you find it ? pleaz

Answer 3

For what x does sin(x)=0 and cos(x)=1?
For what x does cos(x)=0 and sin(x)=-1?

Answer 4

ok thank you

Answer 5

How did you find that sinx = 0 or cosx = 0

Answer 6

If sin(x)=0 and cos(x)=1 then cos^5(x) - sin^5(x)= 1-0=1
If sin(x)=-1 and cos(x)=0 then cos^5(x) - sin^5(x)= 0-(1)^5=0+1=1

Answer 7

These are may be some of the solutions, but if you use a graphing method, then you realize that these are the only one

Answer 8

ok
thanks

Answer 9

I have another problem can u help me

Answer 10

http://www.wolframalpha.com/input/?i=plot+%28Cos%28x%29^5+-+sin%28x%29^5%2C+1%29+for+x+between+-5+Pi+to+5+Pi

Answer 11

\[(a,b,c)\in IR ^{3}: abc=1 prove that: (a-1+\frac{ 1 }{ b })*(b-1+\frac{ 1 }{ c })(c-1+\frac{ 1 }{ a })\le 1\]

Answer 12

Replace c by 1/(ab) in the expression to get
\[
\left(a+\frac{1}{b}-1\right) \left(\frac{1}{a}+c-1\right)
\left(b+\frac{1}{c}-1\right)= \\ \left(a+\frac{1}{b}-1\right)
\left(\frac{1}{a b}+\frac{1}{a}-1\right) (a b+b-1)
\]
Call
\[
f(a,b)=\left(a+\frac{1}{b}-1\right)
\left(\frac{1}{a b}+\frac{1}{a}-1\right) (a b+b-1)\\
\text{ Notice that }
\\
f(1,1)=1
\]

Answer 13

The way a I did it, I found that the gradient of f is zero at (1,1). The Hessian determinant of f at (1,1) is 3>0 and the second derivative with respect to a at (1,1) is -2. Hence f has a maximum at (1,1) which is 1. We are done.
The computations should be done by a machine and that is how I did them and I refuse to do them by hand.