Question 815617
Note: When I first posted a solution, I had neglected to set my calculator to radian mode. So there were errors in the solution. This posting has these errors corrected.<br>
{{{3sin^2(w)-17sin(w)+10=0}}}
First we factor:
{{{(3sin(w)-2)(sin(w)-5)=0}}}
If you cannot see how the equation factored this way, then try using a temporary variable. Let q = sin(w). Then the equation becomes {{{3q^2-17q+10=0}}}. It should not be too difficult to see that it factors into (3q-2)(q-5). Then just replace the q's.<br>
Next we use the Zero Product Property:
3sin(w)-2 = 0 or sin(w)-5=0
Solving 3sin(w)-2 = 0 ...
3sin(w) = 2
sin(w) = 2/3
Using inverse sin we get a reference angle of (approximately) 0.73. Since the 2/3 is positive and since sin is positive in the 1st and 2nd quadrants we get a general solution of:
{{{w = 0.73 + 2pi*n}}} (for 1st quadrant angles)
{{{w = pi-0.73 + 2pi*n = 2.41 + 2pi*n}}} (for 2nd quadrant angles)<br>
Solving sin(w) - 5 - 0 ...
sin(w) = 5
But sin is never larger than 1 (or less than -1). So there is no solution for this.<br>
Last we use the general solution equations to find the specific solutions in the specified interval. 
From {{{w = 0.73 + 2pi*n}}}
if n = 0 then w = 0.73
if n = 1 (or larger) then w is too large for the interval
if n = -1 (or smaller) then w is too small for the interval
From {{{w = 2.41 + 2pi*n}}} (for 2nd quadrant angles)<br>
if n = 0 then w = 2.41
if n = 1 (or larger) then w is too large for the interval
if n = -1 (or smaller) then w is too small for the interval<br>
So the only solutions in the given interval are 0.73 and 2.41.