Video: Solving Trigonometric Equations in Quadratic Form

Find the set of values satisfying 3 sinΒ² πœƒ βˆ’ 2 sin πœƒ cos πœƒ = 0, where 0Β° ≀ πœƒ < 360Β°. Give the answer to the nearest minute.

05:10

Video Transcript

Find the set of values satisfying the equation three sin squared πœƒ minus two sin πœƒ cos πœƒ equals zero, where πœƒ is greater than or equal to zero degrees and less than 360 degrees. Give the answer to the nearest minute.

Here, we have a trigonometric equation. Now, this equation isn’t particularly easy to solve yet, since we have sin πœƒ and cos πœƒ in the same equation. Instead, we notice it’s set equal to zero, and so it looks a little bit like a quadratic. To solve a quadratic equation, we begin by factoring. So, let’s look at how we can factor the expression three sin squared πœƒ minus two sin πœƒ cos πœƒ.

We should spot that both terms, three sin squared πœƒ and negative two sin πœƒ cos πœƒ, have a common factor of sin πœƒ. This means we can factor by taking out this factor of sin πœƒ. We then divide each part by sin πœƒ. Well, three sin squared πœƒ divided by sin πœƒ is three sin πœƒ. Then, negative two sin πœƒ cos πœƒ divided by sin πœƒ is negative two cos πœƒ. So, our equation is now sin πœƒ times three sin πœƒ minus two cos πœƒ equals zero. So, what do we do next?

Well, we recall that for any value of πœƒ, sin πœƒ and three sin πœƒ minus two cos πœƒ are just numbers. And so, for the product of two numbers to be equal to zero, either one or other or both of those numbers must themselves be equal to zero. In other words, either sin πœƒ is equal to zero or three sin πœƒ minus two cos πœƒ is equal to zero. We can solve this first equation by performing inverse operations. We’re going to find the inverse sin of zero, and that gives us zero.

But this isn’t the only solution to the equation sin πœƒ equals zero. Remember, we’re interested in values of πœƒ greater than or equal to zero and less than 360 degrees. And so, we could find the other solution by looking at the curve 𝑦 equals sin of π‘₯. We see it’s equal to zero here, here, and here. The first solution is when πœƒ is equal to zero or zero degrees. The second solution is when πœƒ is equal to 180 degrees. Then, the third is πœƒ is 360 degrees. But actually, we said πœƒ must be less than 360, so we disregard this.

And in fact, instead of using the graph, we could simply recall that sin πœƒ is equal to sin of 180 minus πœƒ. Given one solution of πœƒ, we find the other by subtracting it from 180 degrees. Then, any further solutions are found by adding or subtracting multiples of 360. And this is because the curve is periodic and has a period of 360 degrees. So, we found two solutions to our original equation. But what are we going to do about the equation three sin πœƒ minus two cos πœƒ equals zero?

Well, here, we need to recall a trigonometric identity; that is, tan πœƒ equals sin πœƒ over cos πœƒ. Let’s add two cos πœƒ to both sides of this equation. That gives us three sin πœƒ equals two cos πœƒ. Next, we should spot that if we now divide through by cos πœƒ, we get something that looks a little bit like our expression for tan. We get three sin πœƒ over cos πœƒ equals two, which we can then write as three tan πœƒ equals two. Next, we divide through by three to find an expression for tan πœƒ. So, tan πœƒ is equal to two-thirds.

And we’re now ready to solve like we solved our equation for sin πœƒ. This time, πœƒ is equal to inverse tan of two-thirds or arctan of two-thirds. That gives us πœƒ is 33.69 and so on. By using the degrees, minutes, and seconds button on our calculator, we get 33 degrees, 41 minutes, and 24 seconds. So, correct to the nearest minute, that’s 33 degrees and 41 seconds. Remember, if we don’t have this button on our calculator, what we can do is take the decimal part and multiply that by 60. That gives us 41.4, which is 41 minutes correct to the nearest minute.

To find the other solutions in our interval for πœƒ, we recall that the tangent function is periodic and it has a period of 180 degrees. So, we find another solution by adding 180 degrees to 33.69 and so on. That gives us 213.69 or 213 degrees and 41 minutes. We can use the squiggly brackets to represent the set containing the numbers we require. That’s πœƒ is zero, 33 degrees and 41 minutes, 180 degrees, or 213 degrees and 41 minutes.

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