SOLUTION: solve -2sin^2x=3sinx+1 for exact solutions over the interval [0,2π]
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Question 683200: solve -2sin^2x=3sinx+1 for exact solutions over the interval [0,2π]
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
"exact solutions" is code for: This problem involves special angles and we should put away our calculators.
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 to each side:
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:
After you factor it replace the q's with sin(x)'s and you'll get:
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.
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 has a sin of 1/2. And since sin is negative in the 3rd and 4th quadrants our general solution for this equation is:
for the 3rd quadrant angles
(or ) for the 4th quadrant angles
These simplify to:
for the 3rd quadrant angles
(or ) for the 4th quadrant angles
For the equation sin(x) = -1 we should know that only (and co-terminal angles) will have a sin of -1. So the general solution for this is:
These three general solution equations express the infinite set of angles that fit your equation.
Your problem asks for solutions over the interval [0,2π]. 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 :
If n = 0 then x =
If n = 1 (or other positive integers, x is greater than
If n = -1 (or other negative integers, x is below 0
For the equation
If n = 0 (or any negative integer, x is below 0
If n = 1 then x =
If n = 2 (or larger positive integers, x is greater than
For the equation
If n = 0 then x is below
If n = 1 (or other positive integers, x is greater than
If n = -1 (or other negative integers, x is below 0
So there are only three solutions in the interval [0,2π]: , and
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