Question Video: Solving Trigonometric Equation Involving Special Angles Mathematics

Find the set of values satisfying tan π‘₯ = βˆ’1/√3, where 0 ≀ π‘₯ < 2πœ‹.

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Video Transcript

Find the set of values satisfying tan of π‘₯ equals negative one over root three, where π‘₯ is greater than or equal to zero and less than two πœ‹.

So let’s look at the equation we have, tan of π‘₯ is equal to negative one over root three. One way we have to solve this equation is to perform the inverse operation to tan, that is, to take the inverse tan of both sides. When we do, the left-hand side becomes π‘₯ and the right-hand side becomes the inverse tan of negative one over root three. And what we could then do is type this into a calculator. But actually, the inverse tan of negative one over root three can be calculated by using our special trigonometric values. Let’s remind ourselves how we calculate these by using a table. The values we need to know are sin, cos, and tan of πœ‹ by six, πœ‹ by four, and πœ‹ by three radians.

We fill the first two rows out by writing one, two, three and then three, two, one. We then add a denominator to each of these numbers, and that denominator is two throughout. We then find the square root of the numerator. But of course, the square root of one is simply one, so we don’t write that for one-half. And those are the special values we need to know for sin and cos of πœ‹ by six, πœ‹ by four, and πœ‹ by three radians. tan is not quite straightforward. To find the associated tan values, we divide the sine value by the corresponding cosine value. And because the denominators are the same, this just looks like dividing the numerators. So tan of πœ‹ by six is one over root three. Then tan of πœ‹ by four is root two over two divided by root two over two. Well, that’s just one. Then, tan of πœ‹ by three is root three over two divided by one over two, which is root three over one or root three.

Now, we’re going to compare these values to our equation. We have the inverse tan of negative one over root three. We can see that tan of πœ‹ by six is one over root three. So this must mean that the inverse tan of one over the square root of three is πœ‹ by six. This isn’t quite what we have though. We have negative one over root three. So what next? Well, we can recall that the inverse tangent function is an odd function. This means that the inverse tan of negative π‘₯ is the same as the negative inverse tan of π‘₯. And so our value of π‘₯ must be equal to the negative inverse tan of one over root three. We just saw the inverse tan of one over root three was equal to πœ‹ by six, so π‘₯ must be equal to negative πœ‹ by six.

Sadly, though, we are still not finished. We’re told that π‘₯ takes values greater than or equal to zero and less than two πœ‹. We see that our value of π‘₯ is currently outside of this interval. And so next, we recall that the tangent function is periodic; that is, it repeats, and it does so every πœ‹ radians. So we’re able to find a second value of π‘₯ by adding πœ‹ to the value we have. Negative πœ‹ by six plus πœ‹ is five πœ‹ by six. That is within the interval required. Then we might notice that if we add πœ‹ again, we get 11πœ‹ by six. That’s also within the interval. If, however, we were to add πœ‹ once again, our value would be outside the relevant interval.

And so we have our set of values. Let’s just write them using set notation. And so we see the set of values is the set containing five πœ‹ by six and 11πœ‹ by six.

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