Question 1009489
<pre>
{{{2arcsin(x) + arccos(x) = pi}}}
arcsin(x) means "The angle whose sine is x between {{{-pi/2}}} and {{{pi/2}}} 
arccos(x) means "The angle whose cosine is x" between {{{0}}} and {{{pi}}}
"CoSine" means "Complement's Sine 
We must consider two cases.  
Case 1: when x is positive
Then the arcsin(x) and arcos(x) are both in QI
So x will be  
90°-the angle whose sine is x.
But since we are using radians instead of degrees, it is:
{{{pi/2}}} - the angle whose sine is x. 
So {{{arccos(x)}}}{{{""=""}}}{{{pi/2-arcsin(x)}}}
and therefore
{{{2arcsin(x) + arccos(x)}}}{{{""=""}}}{{{pi}}}
becomes
{{{2arcsin(x) + pi/2-arcsin(x)}}}{{{""=""}}}{{{pi}}}
{{{arcsin(x) + pi/2}}}{{{""=""}}}{{{pi}}}
Multiply through by LCD of 2 to clear the fraction:
{{{2arcsin(x) + pi}}}{{{""=""}}}{{{2pi}}}
{{{arcsin(x) + pi/2}}}{{{""=""}}}{{{pi}}}
{{{arcsin(x)=pi-pi/2}}}
{{{arcsin(x)=pi/2}}}
Therefore the Case 1 solution is x=1 since {{{sin(pi/2)=1}}}

Case 2: when x is negative
 
Then the arcsin(x) is a negative angle in QIV 
And so 2arcsin(x) is an even more negative angle than arcsin(x)

arccos(x) is a positive angle in QII less than {{{pi}}}

So 2arcsin(x)+cos(x) is the sum of a positive angle less than
pi, and a negative angle.  The sum of a positive angle less 
than {{{pi}}} added to a negative angle can never equal to {{{pi}}}, 
so there is no solution to case 2.

Thus the only solution is x=1.

Edwin</pre>