Solve each equation (x in radians and θ in degrees) for all ex...

Question

Solve each equation (x in radians and θ in degrees) for all exact solutions where appropriate. Round approximate answers in radians to four decimal places and approximate answers in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures.
6 sin² θ + sin θ = 1

6 sin² θ + sin θ = 1

Accepted Answer

Hello, today we are going to be solving for the given trigonometric equation. We will be representing our answer in terms of degrees and we'll be rounding to the nearest 10th if necessary. So we are given 10 sine squared data plus three sine data is equal to one. Whenever you want to solve for trigonometric equation you wanna solve for the values of data that satisfy the equation. In order to start this process, we need to set the equation equal to zero. Let's start by subtracting one to both sides of this equation. This will allow us to rewrite it as 10 sine squared theta plus three sign data minus one is equal to zero. We can further simplify this equation by using a substitution. We're going to allow X to equal the sine of the. This will allow us to rewrite the equation as 10 X squared plus three X minus one is equal to zero. Currently, what we have is a factor trinomial. On the left hand side of the equation, the trinomial will factor to leave us with the quantity five X minus one multiplied by the quantity two X plus one equal to zero. Using the zero property, we can set both of these multiples equal to zero. So we will end up with two separate equations. We will have five X minus one equal to zero and we will have two X plus one equal to zero. Now, what we're going to do is we're going to solve for X in both sides of this equation. On the left hand side, we are going to add one to both sides. That will leave us with five X is equal to one and dividing both sides by five will leave us with X is equal to 1/5. On the right hand equation, we will subtract one to both sides that will leave us with two X is equal to negative one and dividing both sides of this equation by two will leave us with X is equal to a negative half. Now, since we are solving for the trigonometric equation, we don't want to solve for the values of X rather we want to solve for the values of theta. So we're going to have to res substitute sine theta for X that will leave us with the equations. Sine theta is equal to 1/5 and sine theta is equal to negative one divided by two. So now what we want to do is you want to use the help of a calculator and a unit circle to help us solve for the values of data, we will label the leftmost equation as one and the rightmost equation as two, let's go ahead and start by solving equation two. So when we look at equation two, we have sine theta is equal to a negative half. In order to solve for this, for this equation, we need to ask ourselves what is an angle theta that we can plug in a sign that would give us the value of negative one divided by two. We recall that any terminal point on the unit circle is represented as cosine comma sine. So in other words, we want to ask ourselves what values on the unit circle is s equal to negative one half. While sine is negative in both the 3rd and 4th quadrant of the unit circle at the angle 210 degrees, we have the terminal point negative square root three divided by two comma negative one divided by two. And in the fourth quadrant, we have the angle 330 degrees which has a terminal point of square root three divided by two comma negative one divided by two at both of these angles. Signed data is equal to a negative half. So there are two solutions for data. We have data is equal to 210 degrees and we have data is equal to 330 degrees. Now let's go ahead and sulfur the values of the in the first equation. Now 1/5 is not a standard value for sign on the unit circle. So we're going to need the help of a calculator. In order to solve for theta, we will need to take the sine inverse of both sides of this equation that will leave us with the is equal to the sine inverse of 1/5. Now, there's one thing you need to remember about sign inverse sine inverses domain is given to us as pi divided by two comma negative pi divided by two. Or in other words, the domain of sine theta sine inverse data is the right half of the unit circle. This is important because when we plug this value into a calculator, we are going to get a value of data as data is equal to 11.5 degrees. 11.5 degrees is a reference angle that exists within this this domain. No, since we were given a positive output for, for sine, we need to find another output for sine theta that is positive as well, sine is positive in both the 1st and 2nd quadrant. So there should be a solution for sine theta that has the same reference angle of 11.5 degrees. In order to find the solution, we're going to set a value of data equal to 180 degrees minus 11.5 degrees. This will give us a value of data equal to 168.5 degrees. So the fourth solution to theta will be located at 168.5 degrees. So currently we have four total solutions for the. Now one thing to note is that these would be the only four solutions if we were given the, the domain from 0 to 2 pi or 0 to 360 degrees. If we take a look back at the original problem, we are not given a restriction of the domain. So what we need to do is we need to represent each value of the, with consideration to any possible co terminal angle. How can we do this? Well, recall that the period for sign is equal to 360 degrees. Now, a co terminal angle can be found by adding any amount of 360 degrees to a single angle. Or in other words, we can rotate about this angle 360 degrees any N amount of times. So we're going to include a multiple N which represents any rotation of the unit circle to each of the values of the. That means we can rewrite the solutions of data to be 11.5 degrees plus 360 N, 168.5 degrees plus 360 N, 210 degrees plus 360 N and 330 degrees plus 360 N. This is also going to be where N is any integer and with that being said, these are going to be the four solutions for the angle theta in the given trigonometric equation. And the overall solution to the problem is going to be B so I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

Key Concepts

Solving Quadratic Trigonometric Equations

General Solutions for Sine Equations

Converting Between Radians and Degrees and Using Least Nonnegative Angles

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