# Lesson Video: Solving a Trigonometric Equation Mathematics

In this video, we will learn how to solve a trigonometric equation using factoring or squaring.

20:49

### Video Transcript

In this video, we will learn how to solve a trigonometric equation using factoring or squaring. The equations that we’ll look at will involve at least one of the trigonometric functions sine, cosine, and tangent. Before looking at these new methods, we will refamiliarize ourselves with solving simple trigonometric equations.

We recall that equations of the form sin 𝜃 equals 𝑘, cos of two 𝜃 equals 𝑘, tan of 𝜃 minus 30 equals 𝑘 can all be solved using either a graph or a CAST diagram. We should also recall some key properties relating to the sine, cosine, and tangent functions. The sin of angle 𝜃 is equal to the sin of 180 degrees minus 𝜃. The cos of 𝜃 degrees is equal to the cos of 360 degrees minus 𝜃. And the tan of any angle 𝜃 is equal to the tan of 180 degrees plus 𝜃.

Before solving a trigonometric equation, it is often helpful to consider how many solutions we expect the equation to have. We can identify the number of times a trigonometric function is equal to a particular value in a given interval by drawing a horizontal line across its graph at this value and then counting the number of times this line intersects the graph. For example, if we want to determine the number of solutions to the equation sin 𝑥 equals 0.5 in the interval where 𝑥 is greater than or equal to zero and less than or equal to 360 degrees, we draw a horizontal line at 𝑦 equals 0.5. As this horizontal line crosses the graph twice in the given interval, we can conclude that the equation sin 𝑥 equals 0.5 has two solutions in the interval 𝑥 is greater than or equal to zero and less than or equal to 360 degrees.

We will now look at a more complex example of this type of problem. Before doing so, we also need to recall one of the key trigonometric identities. This identity defines the relationship between the three trig functions. For any angle 𝜃, the tan of 𝜃 is equal to the sin of 𝜃 divided by the cos of 𝜃. We will now look at an example where we need to use this identity.

If 𝑥 is greater than or equal to zero degrees and less than or equal to 360 degrees, then the number of solutions of the equation four sin 𝑥 is equal to the tan of 𝑥 is what.

This equation involves two trig ratios: sine and tangent. We recall that we can express the tangent function in terms of the sine and cosine functions. The tan of 𝑥 is equal to the sin of 𝑥 over the cos of 𝑥. Substituting this into the right-hand side of our equation, we have four sin 𝑥 is equal to sin 𝑥 over cos 𝑥. Next, we can subtract sin 𝑥 over cos 𝑥 from both sides, giving us four sin 𝑥 minus sin 𝑥 over cos 𝑥 equals zero.

At this stage, we might be tempted to divide the equation through by the shared factor of sin 𝑥. However, doing this could potentially result in the loss of some solutions if the factor we divide by is equal to zero. This means that instead we will factor sin 𝑥 out of the left-hand side of the equation. This gives us sin 𝑥 multiplied by four minus one over cos 𝑥 is equal to zero.

We now have a product that is equal to zero. And the only way a product can equal zero is if at least one of the factors itself equals zero. This means that we need to solve the two equations sin 𝑥 equals zero and four minus one over cos 𝑥 equals zero. Recalling the graph of the sine function as shown, we see that sin 𝑥 is equal to zero three times in the interval where 𝑥 is greater than or equal to zero and less than or equal to 360 degrees. These solutions are zero, 180, and 360 degrees. However, in this question, we are only interested in the number of solutions. sin 𝑥 is equal to zero three times between zero and 360 degrees inclusive.

Let’s now consider the second equation, four minus one over cos 𝑥 equals zero. Multiplying through by cos of 𝑥 gives us four cos of 𝑥 minus one equals zero. We can then add one to both sides such that four cos 𝑥 equals one and, finally, divide through by four such that the cos of 𝑥 is equal to one-quarter.

Recalling the graph of the cosine function and drawing a horizontal line across the graph at 𝑦 equals one-quarter, we find that there are two values of 𝑥 in the interval 𝑥 is greater than or equal to zero and less than or equal to 360 degrees for which the cos of 𝑥 equals one-quarter. Whilst we could calculate these exact values, we don’t need to in this question. However, it is clear that these are not the same values for which sin 𝑥 equals zero, as one of the solutions lies between zero and 90 degrees and the second solution lies between 270 and 360 degrees.

We can therefore conclude that there are five solutions to the equation four sin 𝑥 equals tan 𝑥 between zero and 360 degrees inclusive. The correct answer is five.

We will now look at an example where we need to find all the solutions of a more complex trigonometric equation by factoring.

Find the set of values satisfying tan squared 𝜃 plus tan 𝜃 equals zero, where 𝜃 is greater than or equal to zero degrees and less than 180 degrees.

Upon inspection, we notice that this is a quadratic equation in tan 𝜃. We can begin by factoring the left-hand side of our equation. This gives us tan 𝜃 multiplied by tan 𝜃 plus one is equal to zero. Setting each of these factors equal to zero, we have tan 𝜃 equals zero or tan 𝜃 plus one equals zero. Subtracting one from both sides of our second equation, we now have two solutions: the tan of 𝜃 equals zero or the tan of 𝜃 equals negative one.

Next, we recall the graph of the tangent function as shown. Whilst we’ve drawn it for values of 𝜃 between zero and 360 degrees, it is important to note in this question that we are only looking for solutions greater than or equal to zero and less than 180 degrees. We can see from the graph that the tan of 𝜃 equals zero at zero degrees. This is also true for 180 and 360 degrees. However, these two values are not in the set of values for 𝜃. The equation tan 𝜃 equals zero therefore has one solution where 𝜃 equals zero degrees.

If we draw a horizontal line on our graph at 𝑦 equals negative one, we see that this intersects the graph of 𝑦 equals tan 𝜃 once between zero and 180 degrees. This value lies somewhere between 90 and 180 degrees. Considering the equation tan 𝜃 equals negative one, we can take the inverse tangent of both sides. This gives us 𝜃 is equal to the inverse tan of negative one.

Typing this into our calculator gives us 𝜃 is equal to negative 45 degrees. This value is, however, outside the required interval for 𝜃. Recalling the periodicity of the tangent function, we can find the second solution to the equation by adding 180 degrees. This is because the tan of 𝜃 is equal to the tan of 180 degrees plus 𝜃. We need to add 180 to negative 45. This gives us an answer of 135 degrees, which does lie within the required interval for 𝜃. This is therefore a second valid solution.

From the graph, we can see that there are no further solutions such that 𝜃 is greater than or equal to zero and less than 180 degrees. We can therefore conclude that the set of values that satisfy tan squared 𝜃 plus tan 𝜃 equals zero are zero degrees and 135 degrees.

It is important to note at this stage that a common mistake would be to divide the initial equation by tan 𝜃. This would mean, however, we would lose one of the solutions to the original equation, as it is possible for tan 𝜃 to equal zero. Indeed, this was one of the equations we were subsequently required to solve. We should also take careful note of the interval in which we’re looking for solutions. It is common for the interval to be 𝜃 is greater than or equal to zero and less than or equal to 360 degrees. However, as in this question, this is not always the case. Additional values may well be valid solutions to the equation. But if they are out of the specified interval, then they are not correct in the context of the problem.

Before looking at one final example, we need to consider a second trigonometric identity. This identity is known as the Pythagorean identity. And it states that for all values of 𝜃, sin squared 𝜃 plus cos squared 𝜃 equals one. This identity can be used to solve some specific types of trigonometric equations. In the final example, we will need to use this after squaring both sides of the equation. It is important to note, however, that squaring both sides of an equation can be risky if we do not take great care. This is because squaring and finding the square root are not one-to-one operations. When squaring, we can create an extra solution. Therefore, if we need to solve a trigonometric equation by squaring, we must subsequently check all our solutions in the original equation to ensure we have not obtained any extraneous values.

By first squaring both sides, or otherwise, solve the equation four sin 𝜃 minus four cos 𝜃 is equal to the square root of three, where 𝜃 is greater than zero degrees and less than or equal to 360 degrees. Be careful to remove any extraneous solutions. Give your answers to two decimal places.

The question advises that we approach the problem by first squaring both sides of the equation. Doing so, we obtain four sin 𝜃 minus four cos 𝜃 all squared is equal to root three squared. Distributing the parentheses and then collecting like terms on the left-hand side gives us 16 sin squared 𝜃 minus 32 sin 𝜃 cos 𝜃 plus 16 cos squared 𝜃. On the right-hand side, root three squared is equal to three.

Next, we recall the Pythagorean identity, which states that sin squared 𝜃 plus cos squared 𝜃 is equal to one. Simplifying the left-hand side further gives us 16 multiplied by sin squared 𝜃 plus cos squared 𝜃 minus 32 sin 𝜃 cos 𝜃. Replacing sin squared 𝜃 plus cos squared 𝜃 with one gives us the equation 16 minus 32 sin 𝜃 cos 𝜃 equals three. Subtracting 16 from both sides of this equation gives us negative 32 sin 𝜃 cos 𝜃 is equal to negative 13. We can then divide through by negative 32 such that sin 𝜃 cos 𝜃 equals 13 over 32.

We now have two equations in the two variables sin 𝜃 and cos 𝜃. This means that the system of equations can be solved simultaneously. Adding four cos 𝜃 to both sides of our original equation, we have four sin 𝜃 is equal to root three plus four cos 𝜃. Dividing both sides of this equation by four, we have sin 𝜃 is equal to root three plus four cos 𝜃 all divided by four.

After clearing some space, we will now consider how we can solve these two simultaneous equations. We will begin by substituting the expression for sin 𝜃 in equation two into equation one. This gives us root three plus four cos 𝜃 over four multiplied by cos 𝜃 is equal to 13 over 32. We can simplify this equation by firstly distributing the parentheses. We can then multiply through by 32, giving us eight root three cos 𝜃 plus 32 cos squared 𝜃 equals 13. Finally, subtracting 13 from both sides of this equation, we have the quadratic equation in terms of cos 𝜃 as shown.

This can be solved using the quadratic formula, where 𝑎 is 32, 𝑏 is eight root three, and 𝑐 is negative 13. Substituting in these values and then simplifying gives us cos of 𝜃 is equal to negative three plus or minus the square root of 29 all divided by eight. Taking the inverse cosine of both sides with positive root 29 gives us 𝜃 is equal to 62.829 and so on. To two decimal places, this is equal to 62.83 degrees. Taking the inverse cosine of our equation with negative root 29 gives us 𝜃 is equal to 152.829 and so on. This rounds to 152.83 degrees to two decimal places.

We were asked to give all solutions that are greater than or equal to zero degrees and less than or equal to 360 degrees. We therefore need to consider the symmetry of the cosine function such that the cos of 𝜃 is equal to the cos of 360 degrees minus 𝜃. Subtracting each of our values from 360 degrees gives us further solutions 297.17 degrees and 207.17 degrees to two decimal places.

We have therefore found four possible solutions to the given equation. However, we were reminded in the question to remove any extraneous solutions, these extra solutions that were created when we squared our original equation. We need to substitute each of our four solutions into the initial equation to check they are valid.

The initial equation was four sin 𝜃 minus four cos 𝜃 equals root three. Substituting 𝜃 equals 62.83 into the left side of our equation gives us an answer of root three. This means that this is a valid solution. However, when we substitute 𝜃 is equal to 152.83 degrees into the left-hand side of our equation, we do not get root three. This means that this is not a valid solution. Repeating this process for 207.17 degrees and 297.17 degrees, we see that 207.17 is a valid solution, whereas the fourth answer of 297.17 is not. We can therefore conclude that there are two solutions that satisfy the equation in the given interval of 𝜃, which are 62.83 and 207.17 degrees.

We will now summarize the key points from this video. We saw in this video that some trigonometric equations can be solved by factoring. It is extremely important that any common factors are factored by and not divided by, as this will avoid the potential loss of solutions if these factors are equal to zero. We also saw that some trigonometric equations can be solved by squaring both sides. Whenever we use this approach, it is important to avoid creating any extraneous solutions. We can check for these by substituting in any solutions back in to the original equation.

We saw that the two identities tan of 𝜃 is equal to the sin of 𝜃 divided by the cos of 𝜃 and sin squared 𝜃 plus cos squared 𝜃 equals one can be useful in solving trigonometric equations. Finally, the graphs of trigonometric functions, their properties, and the CAST diagram can be used to determine additional solutions following the determination of a principal angle.

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