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4q = p + 2p n q= Þ p pn + 4 2 n = 0, ±1, ±2,K æa ö Example 8 Solve sin ç ÷ = 0 . è7ø Solution Again, not much to this problem. Using a unit circle it isn’t too hard to see that the solutions to this equation are, a = 0 + 2p n 7 a = p + 2p n 7 a = 14p n n = 0, ±1, ±2,K a = 7p + 14p n Þ This next example has an important point that needs to be understood when solving some trig equations. Example 9 Solve sin ( 3t ) = 2 . Solution This example is designed to remind you of certain properties about sine and cosine.

In that example we noted that -1 £ cos (q ) £ 1 and so the second equation will have no solutions. Therefore, the solutions to the first equation will yield the only solutions to our original equation. 0393 - 2p n n = 0, ±1, ±2,K Note that we did get some negative numbers here and that does seem to violate the general form that we’ve been using in most of these examples. However, in this case the “-” are coming about when we solved for x after computing the inverse cosine in our calculator. There is one more example in this section that we need to work that illustrates another way in which factoring can arise in solving trig equations.

If you followed the work from the first few examples in which we were given intervals you should be able to do any of the remaining examples if given an interval. Also, we will no longer be including sketches of unit circles in the remaining solutions. We are going to assume that you can use the above sketches as guides for sketching unit circles to verify our claims in the following examples. The next three examples don’t require a calculator but are important enough or cause enough problems for students to include in this section in case you run across them and haven’t seen them anywhere else.

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Calculus I by Paul Dawkins

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