### Video Transcript

Find the set of values that satisfy
two sin squared π minus the square root of two sin π minus two equals zero given
that angle π is greater than or equal to 180 degrees and less than 360 degrees.

Now, the first thing we may notice
is that our terms are in the form of a quadratic equation. You may be able to see this more
clearly by defining sin π to be the letter π and rewriting the equation as two π
squared minus the square root of two π minus two equals zero. One of the tools that we have at
our disposal to solve quadratic equations is the quadratic formula. In our case, we can use this to
solve for π or sin π.

The quadratic formula uses π, π,
and π to represent the coefficients to the terms in our equation. In our case π is equal to two, π
is equal to negative square root of two, and π is equal to negative two. We can rewrite the quadratic
formula replacing π with two, π with the negative square root of two, and π with
negative two. Letβs now work on simplifying this
equation.

First, we can cancel out these two
negative symbols to make a positive. Next, we can see that negative
square root of two squared evaluates to the square root of two times the square root
of two. This simplifies to the square root
of four or the integer two. We can then perform the remaining
multiplications and find that the final terms in our square bracket is positive 16
and our denominator is four.

Putting this all together, we find
that our equation simplifies to the following. Letβs now work on the second square
root term on our numerator. Adding our numbers, we find that
this term is equal to the square root of 18. When looking at square roots, it
can be useful to factor out perfect squares. In our case, 18 can be written as
nine times two. And nine is a perfect square of
three times three.

Since powers or exponents are
distributive, we can rewrite this as the square root of nine times the square root
of two. The square root of nine evaluates
to three. So this simplifies to three times
the square root of two. Letβs put this simplified term back
into our equation. We can now see that both terms in
our numerator share a common factor of the square root of two. We can therefore factorize our
numerator to simplify the equation further.

Given that we have a plus or minus
symbol in our equation, we can now write our two separate solutions. In our first solution, we have one
plus three, which evaluates to four. We can then cancel the four in our
numerator and our denominator to find that π equals the square root of two. In our second solution, we have a
one minus three. This evaluates to negative two.

Simplifying this by dividing the
top and bottom half of our fraction by two, we find our second solution is that π
is equal to the negative square root of two over two. Remembering that we defined π as
sin π, we can rewrite this as sin π is equal to the square root of two and sin π
is equal to the negative square root of two over two. As an aside, letβs look at another
way we may have reached this solution by factorizing the original equation.

Taking a familiar approach, we draw
two sets of brackets. We know that the first term in each
of our brackets must multiply together to give the first term in our quadratic. Thatβs two π squared. In this case, itβs fairly easy to
convince ourselves that the correct values are two π and π for our brackets. The second term in each of our
brackets must multiply together to give the last term in our quadratic. Thatβs negative two.

There are a number of ways in which
we can reach negative two by multiplying together two integers. When testing these combinations, we
find that none of them multiply together to give the correct middle term of our
quadratic, negative square root of two. This gives us a clue that we may
need to look at noninteger solutions that multiply together to give negative
two.

In this case, the correct
factorization is two π plus the square root of two times π minus the square root
of two. Now that we have two factors that
multiply together to give zero, we can set each of our sets of brackets to zero to
solve. In the first case, we have two π
plus the square root of two equals zero. To simplify, we can take away the
square root of two from both sides, then we can divide by two. For our second solution, we have π
minus the square root of two equals zero.

To simplify, we can add the square
root of two to both sides. And we have π is equal to the
square root of two. Again, remembering that we defined
π as sin π, we can see that these two solutions are equivalent to the solutions
that we found before. Itβs worth noting that this method
is not necessary if youβre more comfortable using the quadratic formula.

Letβs finally look at finding the
values of π. Here, we have the line π¦ equals
sin π between zero and 360 degrees. In order to find the points where
sin π is equal to the square root of two, we can draw another line on our graph at
π¦ equals the square root of two. At the point where these lines
intersect, π¦ will be equal to sin π, which will be equal to the square root of two
as required by our solution.

We can draw this line more easily
by recognizing that the square root of two is approximately 1.4 to one decimal
place. Now that we have drawn this line on
our graph, we can see that there are no points that the two lines intersect. This means that sin π equals the
square root of two is undefined, and we can ignore this solution for our
question.

Letβs now focus on finding the
other solution, by drawing a line at π¦ equals negative square root of two over
two. By the same logic, the intersection
points of the two lines will give us the solutions to the question. We can see where this line will be
more easily by recognizing that negative square root of two over two is
approximately equal to negative 0.7 to one decimal place. Looking at our line, we can see
that we have two intersection points.

We can better understand these
intersection points by recalling one of the exact trigonometric ratios. Since we know that sin of 45
degrees is equal to the square root of two over two, we can observe this point on
our line. We can also combine this fact with
one of the standard rules of sine which states that sin π is equal to sin of a 180
minus π. Using this rule, we can see that
the value for sin 45 and sin 135 degrees is the same.

We can therefore say that the value
of sin 135 degrees is also the square root of two over two, and this can be observed
on our graph. Let us now imagine shifting our π¦
equals sin π line 180 degrees in the positive π direction; specifically, the
portion of our line that lies between zero and 180 degrees. With this transformation, every
point on the line also moves 180 degrees in the positive π direction. This means that our 45-degree
intersection point would now be at 225 degrees and our 135- degree intersection
point would now be at 315 degrees.

Now we have essentially reached our
solutions. We can see that due to the symmetry
of a sine graph, the two points that we have found share their π value with our two
original intersection points. If we were to perform one more
final transformation and reflect our pink line in the π axis, we would see that our
two green points map onto our two blue intersections. The formal way that we can
represent this is using the following formula: negative sin of π equals sin of π
plus 180 degrees.

Letβs take the 45-degree point as
an example and substitute this in. From this equation, we can see that
the negative of sin of 45 degrees is equal to sin of 225 degrees. Since we know the value of sin of
45 degrees, we can substitute this into our equation, and we see that negative
square root of two over two is indeed sin of 225 degrees. This agrees with the value that we
have found on our graph. And the same logic applies to the
315-degree point.

Finally, we should note that the
question gives us a range for π. And π should be greater than or
equal to 180 degrees and less than 360 degrees. Here, we have drawn the range on
the graph with the solid line at π equals 180 degrees, representing that 180
degrees is included within the range, and a dotted line at π equals 360 degrees,
representing that 360 degrees is not included within the range.

We can clearly see that the values
that we have found for sin π equals negative square root of two over two lie within
the range and, therefore, are valid solutions. The set of values that therefore
satisfies the equation is π equals 225 degrees and π equals 315 degrees.