Video: Solving Trigonometric Equations Involving Special Angles

If 0Β° ≀ πœƒ < 180Β°, find the solution set of √2 sin πœƒ cos πœƒ βˆ’ sin πœƒ = 0.

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Video Transcript

If angle πœƒ is greater than or equal to zero degrees and less than 180 degrees, find the solution set of the square root of two sin πœƒ cos πœƒ minus sin πœƒ equals zero.

The first thing we may notice is that both terms in our equation contain a factor of sin πœƒ. Taking out this factor, we find that sin πœƒ multiplied by the square root of two cos πœƒ minus one is equal to zero. We now have an equation where two factors multiplied together equal zero.

We know that in order for this situation to be true, one of our two factors β€” called 𝐴 and 𝐡 in this simple case β€” must also be equal to zero. We can, therefore, set up two cases for our solutions: one in which sin πœƒ equals zero and another in which root two cos πœƒ minus one equals zero.

Let’s first look at the case of sin πœƒ equals zero. This equation is one of the exact trigonometric ratios that you may be familiar with. Let us also plot this on a graph to help visualise the solution. First, let us observe the range given in the question. The question tells us that πœƒ must be greater than or equal to zero. This has been represented by a solid line on the graph at πœƒ equals zero. The question also tells us that πœƒ must be less than 180 degrees. This has been represented by a dotted line on the graph at πœƒ equals 180.

In order to see our solutions on the graph, let us draw another line at 𝑦 equlas zero and observing where the two lines intersect. In doing this, we will find the points at which 𝑦 is equal to sin πœƒ which is also equal to zero, as required for our solutions. Looking at our graph, we can see intersection points at πœƒ equals zero, πœƒ equals 180, and πœƒ equals 360 degrees.

Of these three points, we can eliminate πœƒ equals 360 degrees as it is clearly outside of our range. We can also eliminate πœƒ equals 180 degrees since our question states that πœƒ can be less than, but not equal to 180. Finally, we can see that πœƒ equals zero degrees is included within our range and is therefore one of the solutions to our question. This agrees with the exact trigonometric ratio that we used earlier.

Although we have found a solution, it is important to remember that the question is not yet complete since we haven’t considered the case where the square root of two cos πœƒ minus one is equal to zero. Let us work on rearranging this into a more useful form.

The first thing we can do is add one to both sides of our equation. The next thing we can do is to divide both sides by the square root of two. Again, looking at this, you may recognize that cos πœƒ equals one over the square root of two is one of the exact trigonometric ratios. Using this knowledge, we can say that πœƒ equals 45 degrees in this case.

As with our previous case, let us plot this on a graph to make sure we aren’t missing any solutions. Here, we have the line 𝑦 equals cos πœƒ. We have also marked the range that we are interested in on the diagram.

Let us now draw the line 𝑦 equals one over the square root of two on the diagram and observe the intersection points with our line. At these intersection points, 𝑦 will equal cos πœƒ, which will also equal one over the square root of two, as required for our case. We may not immediately be able to draw this line. And so we can evaluate that one over the square root of two is approximately equal to 0.71 to two decimal places.

We can now more easily visualise this line on our diagram. Looking at our diagram, we can see two intersection points. However, there is only one point that lies within the required range. Inspecting this point, we can see that this agrees with our solution and that πœƒ is indeed at 45 degrees.

Now that we are happy with the solution to our second case, we can combine this with the solution to our previous case. We can, therefore, say the solution set that satisfies our question is πœƒ equals zero degrees and 45 degrees.

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