Problem with finding multiple solutions to a trigonometric equation

In summary, the problem is that the angles are not equidistant and so you need to use two expressions to represent "all" the possible solutions.
  • #1
TW Cantor
54
1

Homework Statement


Find all solutions to the following equation:

sin((5*x/2)+15)=0.433
where 0 < x < 360

Homework Equations





The Attempt at a Solution


i can find the first value of x that satisfies this equation.

(5*x/2)+15=sin^-1(0.433)
5*x/2=10.6581
5*x=21.2162
x=4.2632

but then i am unsure how to find the other values. i know normally the sine curve repeats itself every 180 degrees but since this is not the case in this question i am unsure as to how to find the other solutions
 
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  • #2
You should know that sin(x)= sin(180- x) and sin(x)= sin(n(360)+ x). Since all of (5x/2)+ 15 is inside the sine, set (5x/2)+ 15 equal to all of those.
 
  • #3
1. (5*x/2)+15=sin^-1(0.433)
2. 5*x/2=10.6581
3. 5*x=21.2162
4. x=4.2632

The problem is in step 3-4.

You can't simply divide 21.2162 by 5 because the whole expression (5*x/2+15) is the angle. When you get the equation into this form:

(5*x/2+15) = arcsin(.422)

You should let the argument of the trig fcn (5*x/2+15) equal angle θ. Now you have something like

θ = arcsin(.422)
θ = 26 degrees (rounded)

Since taking the inverse only works for a certain domain (-90 to 90 deg for arcsin) you know you have one missing solution. Because 26 degrees is positive and in quadrant 1, and sin is essentially the y value about the unit circle, you know that 180-26 is the second angle.

Because those two angles are not equidistant (spelling?) you need to use two expressions to represent "all" the possible solutions given.

26 +/- 360k and (180-26) +/- 360k
(where k is any number to represent how many times you are going around to reach the same angle)

If they were equidistant you could represent the solution set in one expression, for instance, if the angle from an arctan was 45 deg, you have another solution at 315 deg. You could say 45 +/- 180k alone.

Now you can substitute the two solution expressions into the equation you were given.
(5*x/2+15) = 26 +/- 360k
(5*x/2+15) = (180-26) +/- 360k

and solve that to get the expression(s) (there will be two in this case) that capture all of the possible solutions.
 
Last edited:

Related to Problem with finding multiple solutions to a trigonometric equation

What is a trigonometric equation?

A trigonometric equation is an equation that involves trigonometric functions, such as sine, cosine, and tangent, as well as variables and constants. These equations are often used in mathematics and physics to model various real-world phenomena.

Why is it a problem to find multiple solutions to a trigonometric equation?

It is a problem to find multiple solutions to a trigonometric equation because it can create ambiguity and make it difficult to determine the exact solution. This is especially true if the solutions are not expressed in a simplified form.

What is the process for solving a trigonometric equation?

The process for solving a trigonometric equation involves using algebraic techniques to manipulate the equation and isolate the variable. This is often done by applying trigonometric identities and properties. Once the variable is isolated, the solutions can be found by using inverse trigonometric functions or by graphing the equation.

How can I tell if a trigonometric equation has multiple solutions?

If a trigonometric equation has multiple solutions, it means that there is more than one value for the variable that satisfies the equation. This can often be determined by graphing the equation and observing where the graph intersects the x-axis. If there are multiple points of intersection, there are multiple solutions to the equation.

Is there a way to express all the solutions to a trigonometric equation?

Yes, there are multiple ways to express all the solutions to a trigonometric equation. One way is to use the general solution, which involves using a variable, such as n, to represent all the possible solutions. Another way is to use the principal value, which is the smallest positive solution to the equation. Additionally, solutions can be expressed in radians or degrees, depending on the context of the problem.

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