### Video Transcript

In this video, we will learn to solve trigonometric equations using the double-angle identity. We will find solutions given a particular range in degrees or radians. The equations that we are dealing with will be able to be simplified using the double-angle identities or the half-angle identities. We will begin by recalling how we can solve simple equations using the CAST method or trig graphs.

As already mentioned, two units for measuring angles are degrees and radians. We recall that 180 degrees is equal to 𝜋 radians. This means that 90 degrees is equal to 𝜋 over two radians and 360 degrees is equal to two 𝜋 radians. We can find the solutions to a trig equation by drawing the appropriate graph or by using the CAST diagram. Whilst there are an infinite number of solutions to our equations, in this video, we will focus on solutions between zero and 360 degrees or zero and two 𝜋 radians. We measure the angles in a counterclockwise direction from the positive 𝑥-axis as shown. We can replace the values in degrees by the corresponding values in radians.

In the first quadrant, labeled A, the values of sin 𝜃, cos of 𝜃, and tan of 𝜃 are all positive. In the second quadrant, labeled S, the value of sin 𝜃 is positive, whereas the value of cos 𝜃 and tan 𝜃 are negative. In the third quadrant, the tan of 𝜃 is positive, whereas the sin of 𝜃 and cos of 𝜃 are negative. This is therefore true when 𝜃 lies between 180 and 270 degrees. Finally, when 𝜃 lies between 270 and 360 degrees, known as the fourth quadrant, the value of cos 𝜃 is positive, whereas the values of sin 𝜃 and tan 𝜃 are negative. We can use this information to solve simple trig equations.

Find the set of all possible solutions of cos 𝜃 equals one-half, given that 𝜃 exists between zero and 360 degrees inclusive.

The square brackets here mean that 𝜃 must be greater than or equal to zero degrees and less than or equal to 360 degrees. If we had curved brackets or parentheses, then 𝜃 would be strictly greater than or less than the value. To solve the equation cos 𝜃 equals one-half, we take the inverse cosine of both sides of the equation. Solving this gives us a value of 𝜃 equal to 60 degrees. At this stage, it is worth sketching our CAST diagram, as we can then see which quadrants our solutions lie in. As the cos of 𝜃 is positive, there will be solutions in the first and fourth quadrants.

We already have the solution in the first quadrant. This is equal to 60 degrees. We can then calculate the solution in the fourth quadrant by subtracting 60 degrees from 360 degrees. This gives us an answer of 300 degrees. The set of solutions of cos 𝜃 equals one-half between zero and 360 degrees are 60 degrees and 300 degrees.

We will now consider some more complicated problems using the double-angle identities.

Our double-angle identities are as follows. Firstly, the sin of two 𝜃 is equal to two sin 𝜃 cos 𝜃. Note that identities are also sometimes written with a triple line, which shows they are true for all values of the variable. The cos of two 𝜃 is equal to cos squared 𝜃 minus sin squared 𝜃. Using the fact that sin squared 𝜃 plus cos squared 𝜃 equals one, cos two 𝜃 is also equal to two cos squared 𝜃 minus one and one minus two sin squared 𝜃. Finally, we have the tan of two 𝜃 is equal to two tan 𝜃 divided by one minus tan squared 𝜃.

In this video, we will not prove these identities. We will simply quote them to solve our equations. In our next question, we will use the fact that sin two 𝜃 is equal to two sin 𝜃 cos 𝜃.

Find the set of possible solutions of two sin 𝜃 cos 𝜃 equals zero, given 𝜃 is greater than or equal to zero and less than 360 degrees.

The square bracket tells us that 𝜃 is greater than or equal to zero degrees, whereas the curved bracket tells us that 𝜃 is strictly less than 360 degrees. We can solve our equation using our knowledge of the double-angle identities. We know that sin of two 𝜃 is equal to two sin 𝜃 cos 𝜃. This means that we need to solve sin of two 𝜃 is equal to zero. Taking the inverse sine of both sides of this equation, we get two 𝜃 is equal to the inverse sin of zero.

Whilst solving this equation would give us one solution, it is worth sketching the graphs of sin 𝜃 and sin two 𝜃 before proceeding. We know that sin 𝜃 is a periodic function, and we are interested in values between zero and 360 degrees. It has a maximum value of one and a minimum value of negative one, as shown. The graph of sin two 𝜃 will be a dilation or enlargement. The graph will be stretched by a scale factor of one-half in the horizontal direction. This means that it will look as shown.

This has a maximum value of one when 𝜃 is 45 degrees and a minimum value of negative one when 𝜃 is 135 degrees. We are interested in the points at which the graph is equal to zero. There are five such points on our graph, at zero, 90, 180, 270, and 360 degrees. We recall though that 𝜃 must be less than 360 degrees. So our last point is not a solution. The four solutions of the equation two sin 𝜃 cos 𝜃 equals zero, where 𝜃 is greater than or equal to zero degrees and less than 360 degrees, are zero, 90, 180, and 270 degrees.

In our next example, we will use one of the double-angle identities involving cos of two 𝜃.

Find the solution set for 𝑥 given the cos of two 𝑥 plus 13 root three multiplied by the cos of 𝑥 is equal to negative 19, where 𝑥 exists between zero and two 𝜋.

The parentheses or curved brackets tell us that 𝑥 is strictly greater than zero and less than two 𝜋. This also indicates that we need to give our solutions in radians. We can solve the equation in this range by using our double-angle identities.

One of these identities states that the cos of two 𝑥 is equal to two cos squared 𝑥 minus one. We can therefore rewrite our equation as shown. We can then add 19 to both sides of the equation, giving us two cos squared 𝑥 plus 13 root three cos 𝑥 plus 18 is equal to zero. If we let 𝑦 equal cos of 𝑥, this can be rewritten as two 𝑦 squared plus 13 root three 𝑦 plus 18 equals zero.

We now have a quadratic equation in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, which we can solve using the quadratic formula as shown. Our values of 𝑎, 𝑏, and 𝑐 are two, 13 root three, and 18, respectively. Substituting these values into the formula, we get two values for 𝑦. Either 𝑦 is equal to negative root three over two or negative six root three. This means that the cos of 𝑥 must be equal to negative root three over two or negative six root three.

As the value of the cos of 𝑥 must lie between negative one and one inclusive, the second option has no solutions. We can now solve the equation the cos of 𝑥 is equal to negative root three over two using our CAST diagram. As the cos of our angle is negative, we will have solutions in the second and third quadrants. One of our solutions will be between 𝜋 over two and 𝜋 and our second between 𝜋 and three 𝜋 over two.

We know that the cos of 30 degrees or 𝜋 over six radians is equal to root three over two. This means that our two solutions will be equal to 𝜋 minus 𝜋 over six and 𝜋 plus 𝜋 over six. This gives us two solutions: five 𝜋 over six and seven 𝜋 over six. The solution set of the equation cos two 𝑥 plus 13 root three cos 𝑥 equals negative 19 are five 𝜋 over six and seven 𝜋 over six.

Before answering our final question, we will consider the half-angle identities.

Firstly, we have the sin of 𝜃 over two is equal to the positive or negative of the square root of one minus cos 𝜃 all divided by two. The second identity is very similar. The cos of 𝜃 over two is equal to the positive or negative square root of one plus cos 𝜃 all divided by two. Finally, we have the tan of 𝜃 over two is equal to the sin of 𝜃 divided by one plus the cos of 𝜃. Alternatively, this is equal to one minus the cos of 𝜃 divided by the sin of 𝜃. We will now consider a question where we need to use one of these identities.

Solve tan of 𝑥 over two is equal to sin 𝑥, where 𝑥 is greater than or equal to zero and less than two 𝜋 radians.

We are told in the question that our solutions must be less than two 𝜋 and greater than or equal to zero. This means that we will be able to solve the equations using the CAST diagram or by sketching our trig graphs. Using one of our half-angle identities, we know that tan of 𝑥 over two is equal to one minus cos 𝑥 divided by sin 𝑥. This means that we can rewrite our equation as one minus cos 𝑥 divided by sin 𝑥 is equal to sin 𝑥

Multiplying both sides of the equation by sin 𝑥 gives us one minus cos 𝑥 is equal to sin squared 𝑥. As sin squared 𝑥 is equal to one minus cos squared 𝑥, we have one minus cos 𝑥 is equal to one minus cos squared 𝑥. We can then add cos squared 𝑥 and subtract one from both sides of our equation. This gives us cos squared 𝑥 minus cos 𝑥 is equal to zero. Factoring out cos 𝑥 gives us cos 𝑥 multiplied by cos 𝑥 minus one is equal to zero. This means that either cos of 𝑥 equals zero or cos of 𝑥 minus one equals zero. The second equation can be rewritten as the cos of 𝑥 equals one.

We now have two equations that can be solved either by using the CAST diagram or by drawing the graph of cos 𝑥. The cosine graph between zero and two 𝜋 looks as shown. Remembering that our values need to be greater than or equal to zero and less than two 𝜋, we see that the graph is equal to zero when 𝑥 is equal to 𝜋 over two and three 𝜋 over two. cos of 𝑥 is equal to one when 𝑥 is equal to zero or two 𝜋. However, two 𝜋 is not within the range of values for 𝑥. This means that there are three solutions to our equation. The tan of 𝑥 over two is equal to the sin of 𝑥 when 𝑥 is equal to zero, 𝜋 over two, and three 𝜋 over two.

We will now summarize the key points from this video. We saw in this video that we can solve trig equations by using the double-angle identities. We can also use adaptions of these equations known as the half-angle identities. In both of these cases, we will usually be given a range of values often between zero and 360 degrees or zero and two 𝜋 radians. We can then use our CAST diagram or the trig graphs to help us solve the equations.