Video Transcript
Find the set of values satisfying
sin of two 𝑥 plus 𝜋 over three equals root two over two, where 𝑥 is greater than
or equal to zero and less than two 𝜋.
We’re going to solve a simple
trigonometric equation in sine, where the solutions are going to be in radians. Before we do, observe the argument
of the trigonometric expression on the left-hand side of our equation. The argument is two 𝑥 plus 𝜋 over
three. In order to aid us later on in this
problem, we’re going to begin by changing the range in which our values could
lie. That is, we’re told 𝑥 is greater
than or equal to zero and less than two 𝜋. We want to make the middle value in
this inequality look exactly like our argument. To achieve this, we’ll begin by
multiplying through by two, giving us an inequality which is equivalent to the
original of two 𝑥 is greater than or equal to zero and less than four 𝜋.
Next, we’ll add 𝜋 by three to each
part of this inequality. So we get a third equivalent
inequality that says that two 𝑥 plus 𝜋 by three is greater than or equal to 𝜋 by
three and less than 13𝜋 by three. Now this is useful because it
allows us to solve the first part of the equation as we would solve any other
trigonometric equation. That is, we perform the inverse
operation to sine. We find the inverse sine of both
sides. The left-hand side of our equation
then simply becomes two 𝑥 plus 𝜋 by three. And then the right-hand side
becomes the inverse sin of root two over two. However, the inverse sin of root
two over two is 𝜋 by four. So we can say that two 𝑥 plus 𝜋
by three is equal to 𝜋 by four.
Now, before we go any further, we
must remember that in the interval zero to two 𝜋, there must be two solutions to
this equation. If the first is 𝜋 by four, the
second is found by subtracting 𝜋 by four from 𝜋. And that’s simply due to the shape
of the curve. And we see it intersects the line
𝑦 equals root two over two in exactly two places. 𝜋 minus 𝜋 over four is three 𝜋
over four. So those are the two solutions to
this equation in the interval zero to two 𝜋. But of course, the sine function is
periodic. And this is where changing the
range over which our solutions can lie is really useful. We can add and subtract multiples
of two 𝜋 to our solutions till we have every solution in the interval 𝜋 by three
to 13𝜋 by three.
Let’s begin by adding two 𝜋 to 𝜋
by four. That gives us nine 𝜋 by four. Then, we add two 𝜋 to three 𝜋
over four, giving us 11𝜋 over four. Now at this point, we know that 11
divided by four is a little under three, whilst 13 divided by three is a little over
four. So we need to continue finding
solutions. Let’s add another multiple of two
𝜋, and we can do this by adding two 𝜋 to the solutions we just found. Nine 𝜋 by four plus two 𝜋 is 17𝜋
by four, and 11𝜋 by four add two 𝜋 is 19𝜋 by four. 19 divided by four is 4.75, and
that’s greater than 13 divided by three. This means we disregard this
solution. It’s outside of the interval we
specified.
17 over four is less than 13 over
three, however, and so this solution is within the interval. Similarly, 𝜋 by four is in fact
less than 𝜋 by three. So this is outside of the range of
solutions. So we see that two 𝑥 plus 𝜋 by
three could be equal to three 𝜋 by four, nine 𝜋 by four, 11𝜋 by four, or 17𝜋 by
four. In order to find the values of 𝑥
that satisfy these four independent equations, we’ll need to solve by subtracting 𝜋
by three and then dividing by two.
When we do that to three 𝜋 by
four, we’re left with five 𝜋 by 24. Similarly, nine 𝜋 by four minus 𝜋
by three and then divided by two is 23𝜋 over 24. Repeating this with 11𝜋 by four
and 17𝜋 by four, that is, subtracting 𝜋 by three and then dividing by two, gives
us 29𝜋 over 24 and 47𝜋 over 24. We can now observe that each of
these solutions is within the original range 𝑥 is greater than or equal to zero and
less than two 𝜋. Let’s write this then in set
notation. The set of values that satisfy our
equation is the set containing five 𝜋 over 24, 23𝜋 over 24, 29𝜋 over 24, and 47𝜋
over 24.
