Question 1030290
<pre><b>
There are infinitely many solutions:

{{{n*expr(pi/2)}}} where n is any integer positive negative or 0. 

{{{sin^8(x)+cos^6(x)=1}}}

{{{(sin^2(x))^4+(cos^2(x))^3=1}}}

{{{(sin^2(x))^4+(1-sin^2(x))^3=1}}}

Let {{{u=sin^2(x)}}}

{{{u^4+(1-u)^3=1}}}

{{{u^4+(1-u)(1-u)(1-u)=1}}}

{{{u^4+(1-u)(1-2u+u^2)=1}}}

{{{u^4+1-2u+u^2-u+2u^2-u^3=1}}}

{{{u^4-u^3+u^2+2u^2-2u-u+1=1}}}

{{{u^4-u^3-u^2+u=0}}}

{{{u(u^3-u^2-u+1)=0}}}

It's easy to see that 1 is a solution
to the cubic polynomial in parentheses.
We use synthetic division

1|1 -1 -1  1
 |<u>   1  0 -1</u>
  1  0 -1  0

So we have factored to above as

{{{u(u-1)(u^2-1)=0}}}

And we can factor once more as

{{{u(u-1)(u-1)(u+1)=0}}}

That has solutions

u=0, u=1, u=-1

Since {{{u=sin^2(x)}}}

{{{sin^2(x)=0}}}, {{{sin^2(x)=1}}}, {{{cross(sin^2(x)=-1)}}}

We eliminate the third case because it will not give 
a real solution.

{{{sin(x)=0}}}, {{{sin(x)="" +- 1}}}

The first gives any multiple of <font face="symbol">p</font> or n<font face="symbol">p</font>,
where n is any integer, positive negative or zero.

The second gives ±<font face="symbol">p</font>/2 plus any multiple of 2<font face="symbol">p</font>,

or <font face="symbol">p</font>/2 + 2k<font face="symbol">p</font> = (1+2k)<font face="symbol">p</font>/2.
where k is any integer, positive negative or zero. 

That's all odd multiples of <font face="symbol">p</font>/2  

Since all multiples of <font face="symbol">p</font> are also multiples
of ±<font face="symbol">p</font>/2, all solutions are the multiples of
<font face="symbol">p</font>/2 

n<font face="symbol">p</font>/2                       <--- answer
(where n is any integer, positive negative or zero.)

So the number of solutions is INFINITY!

Edwin</pre></b>