# Video: Solving Trigonometric Equations in Quadratic Form Involving Special Angles

Darius Garewal

Find the set of values that satisfy 2 sin² 𝜃 − √(2) sin 𝜃 − 2 = 0 given 180° ≤ 𝜃< 360°.

11:20

### Video Transcript

Find the set of values that satisfy two sin squared 𝜃 minus the square root of two sin 𝜃 minus two equals zero given that angle 𝜃 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 𝜃 to be the letter 𝑠 and rewriting the equation as two 𝑠 squared minus the square root of two 𝑠 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 𝑠 or sin 𝜃.

The quadratic formula uses 𝑎, 𝑏, and 𝑐 to represent the coefficients to the terms in our equation. In our case 𝑎 is equal to two, 𝑏 is equal to negative square root of two, and 𝑐 is equal to negative two. We can rewrite the quadratic formula replacing 𝑎 with two, 𝑏 with the negative square root of two, and 𝑐 with negative two. Let’s 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. Let’s 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. Let’s 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 𝑠 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 𝑠 is equal to the negative square root of two over two. Remembering that we defined 𝑠 as sin 𝜃, we can rewrite this as sin 𝜃 is equal to the square root of two and sin 𝜃 is equal to the negative square root of two over two. As an aside, let’s 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. That’s two 𝑠 squared. In this case, it’s fairly easy to convince ourselves that the correct values are two 𝑠 and 𝑠 for our brackets. The second term in each of our brackets must multiply together to give the last term in our quadratic. That’s 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 𝑠 plus the square root of two times 𝑠 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 𝑠 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 𝑠 minus the square root of two equals zero.

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

Let’s finally look at finding the values of 𝜃. Here, we have the line 𝑦 equals sin 𝜃 between zero and 360 degrees. In order to find the points where sin 𝜃 is equal to the square root of two, we can draw another line on our graph at 𝑦 equals the square root of two. At the point where these lines intersect, 𝑦 will be equal to sin 𝜃, 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 𝜃 equals the square root of two is undefined, and we can ignore this solution for our question.

Let’s now focus on finding the other solution, by drawing a line at 𝑦 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 𝜃 is equal to sin of a 180 minus 𝜃. 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 𝑦 equals sin 𝜃 line 180 degrees in the positive 𝜃 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 𝜃 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 𝜃 value with our two original intersection points. If we were to perform one more final transformation and reflect our pink line in the 𝜃 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 𝜃 equals sin of 𝜃 plus 180 degrees.

Let’s 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 𝜃. And 𝜃 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 𝜃 equals 180 degrees, representing that 180 degrees is included within the range, and a dotted line at 𝜃 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 𝜃 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 𝜃 equals 225 degrees and 𝜃 equals 315 degrees.