SOLUTION: solve 3sin^2(w)-17sin(w)+10=0 for all solutions where w is in radians between 0 and 2pi

Question 815617: solve 3sin^2(w)-17sin(w)+10=0 for all solutions where w is in radians between 0 and 2pi
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
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.

First we factor:

If you cannot see how the equation factored this way, then try using a temporary variable. Let q = sin(w). Then the equation becomes . It should not be too difficult to see that it factors into (3q-2)(q-5). Then just replace the q's.

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:
(for 1st quadrant angles)
(for 2nd quadrant angles)

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.

Last we use the general solution equations to find the specific solutions in the specified interval.
From
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 (for 2nd quadrant angles)

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

So the only solutions in the given interval are 0.73 and 2.41.
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