Question 683200
{{{-2sin^2(x)=3sin(x)+1}}}
"exact solutions" is code for: This problem involves special angles and we should put away our calculators.<br>
First we want to transform the equation into one or more equations of the form:
TrigFunction(expression) = number
This equation is in quadratic form for sin(x). So we'll start by getting one side to be zero. Adding {{{2sin^2(x)}}} to each side:
{{{0 = 2sin^2(x)+3sin(x)+1 }}}
Now we factor. This factors fairly easily. If you have trouble seeing this then use a temporary variable:
Let q = sin(x). Then the equation becomes:
{{{0 = 2q^2+3q+1}}}
After you factor it replace the q's with sin(x)'s and you'll get:<br>
{{{0 = (2sin(x)+1)(sin(x)+1) }}}
From the Zero Product Property:
2sin(x) + 1 = 0 or sin(x) + 1 = 0
Solving these for sin(x) we get:
sin(x) = -1/2 or sin(x) = -1
These equations are in the desired form.<br>
Next we find the general solution. As anticipated we have special angle values for sin in both equations. For sin(x) = -1/2 we should recognize the reference angle of {{{pi/6}}} has a sin of 1/2. And since sin is negative in the 3rd and 4th quadrants our general solution for this equation is:
{{{x = pi+pi/6+2pi*n}}} for the 3rd quadrant angles
{{{x = -pi/6+2pi*n}}} (or {{{2pi-pi/6+2pi*n}}}) for the 4th quadrant angles
These simplify to:
{{{x = 7pi/6+2pi*n}}} for the 3rd quadrant angles
{{{x = -pi/6+2pi*n}}} (or {{{11pi/6+2pi*n}}}) for the 4th quadrant angles<br>
For the equation sin(x) = -1 we should know that only {{{3pi/2}}} (and co-terminal angles) will have a sin of -1. So the general solution for this is:
{{{x = 3pi/2+2pi*n}}}
These three general solution equations express the infinite set of angles that fit your equation.<br>
Your problem asks for solutions over the interval [0,2&#960;]. For this we use the general solution equations and replace the n's with integers until we find all the x's in the given interval.
For the equation {{{x = 7pi/6+2pi*n}}}:
If n = 0 then x = {{{7pi/6}}}
If n = 1 (or other positive integers, x is greater than {{{2pi}}}
If n = -1 (or other negative integers, x is below 0
For the equation {{{x = -pi/6+2pi*n}}}
If n = 0 (or any negative integer, x is below 0
If n = 1 then x = {{{11pi/6}}}
If n = 2 (or larger positive integers, x is greater than {{{2pi}}}
For the equation {{{x = 3pi/2+2pi*n}}}
If n = 0 then x is below {{{3pi/2}}}
If n = 1 (or other positive integers, x is greater than {{{2pi}}}
If n = -1 (or other negative integers, x is below 0
So there are only three solutions in the interval [0,2&#960;]: {{{7pi/6}}}, {{{11pi/6}}} and {{{3pi/2}}}