Find the set of values satisfying the equation 97 sin 𝜃 plus 60 cos 𝜃 equals zero, where 𝜃 is greater than zero and less than 360 degrees. Give the answers to the nearest second.
Here we have a trigonometric equation. Now, we have a little bit of a problem. It’s quite difficult to solve trigonometric equations when we’ve got different trigonometric functions. Instead, we spot we have an expression with both sin 𝜃 and cos 𝜃 in it. And we know that there’s an identity that links these. It’s tan 𝜃 equals sin 𝜃 over cos 𝜃. And so, what we’re going to look to do with our equation is rearrange it to find the expression sin 𝜃 over cos 𝜃.
Let’s begin by subtracting 60 cos 𝜃 from both sides. When we do, we get 97 sin 𝜃 equals negative 60 cos 𝜃. Next, we should see that if we divide through by cos 𝜃 on the left-hand side, we’re going to be dividing some multiple of sin 𝜃 by cos 𝜃. That’s the form we want. And so, dividing through by cos 𝜃, we get 97 sin 𝜃 over cos 𝜃 equals negative 60. Now, in fact, we just want sin 𝜃 over cos 𝜃 rather than some multiple of that. So, we’re going to divide through by 97 next. And that gives us sin 𝜃 over cos 𝜃 equals negative 60 over 97 or, using our identity, tan 𝜃 equals negative 60 over 97.
So, how do we solve this equation? Well, we need to perform the inverse operation. That is, we’re going to find the inverse tan of negative 60 over 97. And when we do, we find that one of our solutions is 𝜃 equals negative 31.739 and so on degrees. The problem we have here is that our value of 𝜃 is outside of the interval 𝜃 is greater than zero and less than 360. And so, we have a couple of options here. Let’s begin by considering the shape of the graph that represents the function 𝑦 equals tan of 𝑥. The graph looks a little something like this. We have these vertical asymptotes at 90 degrees and then every 180 degrees either side of that.
Let’s add in the line 𝑦 equals negative 60 over 97 since we’re solving the equation tan 𝜃 is equal to this value. The first solution we have got is negative 31.7. That’s this solution here. But if we look carefully, we notice that there are further solutions along the curve. There’s one somewhere between 90 and 180 and another one somewhere between 270 and 360. In fact, since the tangent function is periodic, that is, it repeats, and it does so every 180 degrees, our solutions occur every 180 degrees. In fact, we can say that for any value of 𝜃, tan of 𝜃 will be the same as tan of 𝜃 plus 180 degrees. And so, a value of 𝜃 that will be inside the interval we’re looking for will be negative 31.7 plus 180, which is 148.26 — well, this is in degrees.
Now, I’ve not rounded. And this is because we’re told to give our answers to the nearest second. We could use the button that represents degrees, minutes, and seconds. Alternatively, we take the decimal number and we multiply it by 60. That gives us 15.65 and so on. That tells us the number of minutes. We have 15 minutes. If we take the decimal part again and multiply that by 60, that tells us the number of seconds. 0.65 times 60 is 39.09 and so on. And that’s in seconds. And so, correct to the nearest second, one of our solutions for 𝜃 is 148 degrees, 15 minutes, and 39 seconds.
And what about the other solution? Well, we’ll add 180 once again. This gives us our second solution in the interval to be 𝜃 equals 328.26 or 328 degrees, 15 minutes, and 39 seconds. If we were to add another 180 degrees, we’d find a value of 𝜃 which is outside of the interval. So, we’re done. And we can use these sort of squiggly brackets to represent the set. The set of values satisfying our equation in the given interval is shown.