Question 1161852
<pre>
arccos(2-3x) = 2arcsin(x)

Take the cosine of both sides:

{{{cos(arccos(2-3x)^"")= cos(2arcsin(x)^"")}}}

The left side is just 2-3x.  For the right side let:
                              
                                   arcsin(x) = &theta;
                                      sin(&theta;) = x 
                                     
{{{2-3x=cos(2theta)}}}
{{{2-3x=1-2sin^2(theta)}}}
{{{2-3x=1-2x^2)}}}
{{{2x^2-3x+1=0}}}
{{{(2x-1)(x-1)=0}}}

Two solutions: 1/2 and 1

Edwin</pre>