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Question Video: Finding the General Solution of the Trigonometric Equation cos (3π‘₯) = sin (π‘₯/4) Mathematics • 10th Grade

Find the general solution to the equation the cos 3π‘₯ = sin (π‘₯/4).

06:23

Video Transcript

Find the general solution to the equation the cos of three π‘₯ is equal to the sin of π‘₯ over four.

We’re given the trigonometric equation cos three π‘₯ is equal to sin π‘₯ over four and asked to find the general solution. So we want to solve this for all possible π‘₯-values that satisfy the given equation. Now we might try using some trigonometric identities to rewrite our left-hand side, which we can do if we write three π‘₯ as two π‘₯ plus π‘₯. We might then try using the sum identity for cos 𝐴 plus 𝐡, which tells us that cos 𝐴 plus 𝐡 is equal to cos 𝐴 cos 𝐡 minus sin 𝐴 sin 𝐡 and with 𝐴 equals two π‘₯ and 𝐡 equal to π‘₯ gives cos two π‘₯ times cos π‘₯ minus sin two π‘₯ times sin π‘₯ equals sin π‘₯ over four.

We’re trying to get the left- and right-hand sides into similar forms, so we might then try using the double angle identities for cos two 𝐴 and sin two 𝐴. This gives, for example, cos squared π‘₯ minus sin squared π‘₯ times cos π‘₯ minus two sin squared π‘₯ times cos π‘₯ equals sin π‘₯ over four. Then, using the Pythagorean identity cos squared 𝐴 plus sin squared 𝐴 equals one gives cos π‘₯ multiplied by four cos squared π‘₯ minus three is equal to sin π‘₯ over four.

We’ve reduced the argument of the cosines on the left to just π‘₯, but on our right-hand side, we still have sin π‘₯ over four. And without somehow manipulating the argument π‘₯ over four into just π‘₯, we’re now nearer to finding solutions for π‘₯. So our method of using the usual trigonometric identities doesn’t seem to be working here. So let’s try looking at this from another angle. If we consider instead part of the graphs of the cosine and sine functions, we see, for example, that cos of zero equals sin of πœ‹ by two, which is equal to one. cos of πœ‹ by two equals sin of πœ‹, which is equal to zero. Similarly, cos of negative πœ‹ by two equals sin of zero equals zero. And cos of negative πœ‹ equals sin of negative πœ‹ by two equals negative one and so on.

In fact, the pattern here is that the cos of an angle πœƒ is equal to the sin of πœ‹ by two plus or minus πœƒ. So, for example, if πœƒ equals πœ‹, taking the positive option, we have cos πœ‹ equal to the sin of πœ‹ by two plus πœ‹, which is the sin of three πœ‹ by two. And the sin of three πœ‹ by two is negative one, which is the same as the cos of πœ‹. Similarly, for the negative option, we have cos πœ‹ equal to the sin of πœ‹ by two minus πœ‹. That’s the sin of negative πœ‹ by two, which again is negative one. And this works for any angle πœƒ.

So now making some space and applying this to the left-hand side of our equation, the cos of three π‘₯, so that πœƒ is three π‘₯, we have the cos of three π‘₯ equal to sin πœ‹ by two plus or minus three π‘₯. And this must equal the right-hand side of our original equation; that’s sin π‘₯ over four. So we now have the sin of πœ‹ by two plus or minus three π‘₯ is equal to the sin of π‘₯ over four, and everything is in terms of sines. Let’s now call πœ‹ by two plus or minus three π‘₯ 𝐴 and π‘₯ over four 𝐡, so that our equation is sin 𝐴 is equal to sin 𝐡.

We can rearrange this to give sin 𝐴 minus sin 𝐡 is equal to zero. And if we can solve this equation, we should be able to find our general solution for π‘₯. To do this, we recall the sum and difference trigonometric identity. This tells us that two times the cos of 𝐴 plus 𝐡 over two multiplied by the sin of 𝐴 minus 𝐡 over two is equal to sin 𝐴 minus sin 𝐡. Now, in our case, sin 𝐴 minus sin 𝐡 is equal to zero, and in our identity, there are two ways that this can occur. Either cos 𝐴 plus 𝐡 over two must be equal to zero or the sin of 𝐴 minus 𝐡 over two must be equal to zero.

Taking the first of these, cos of 𝐴 plus 𝐡 over two equals zero, if we look again at the graph of the cosine function, we see that cos πœƒ is equal to zero when πœƒ is equal to πœ‹ over two plus π‘›πœ‹, where 𝑛 is an integer. This means in our case that 𝐴 plus 𝐡 over two must equal πœ‹ over two plus π‘›πœ‹. And if we multiply through by two, this gives 𝐴 plus 𝐡 is equal to πœ‹ plus two π‘›πœ‹. Making a little space, we can substitute our expressions for 𝐴 and 𝐡. Now subtracting πœ‹ over two from both sides, we have positive or negative three π‘₯ plus π‘₯ over four is equal to πœ‹ plus two π‘›πœ‹ minus πœ‹ over two.

Now, putting our left-hand side over a common denominator of four and simplifying our right-hand side a little, then multiplying both sides by four, we have two possibilities on the left-hand side: either taking first positive 12π‘₯ plus π‘₯, so giving 13π‘₯ is equal to two πœ‹ plus eight π‘›πœ‹, or with negative 12π‘₯ plus π‘₯, we have negative 11π‘₯ is equal to two πœ‹ plus eight π‘›πœ‹. So now clearing a little space, dividing both sides by 13 in our first solution gives us π‘₯ equal to two πœ‹ over 13 plus eight π‘›πœ‹ over 13. Next, dividing our second equation by negative 11, we have π‘₯ equal to two πœ‹ over negative 11 plus eight π‘›πœ‹ over negative 11. And since 𝑛 is any integer, positive or negative, we can rewrite this as negative two πœ‹ over 11 plus eight π‘›πœ‹ over 11.

The general solution to the equation cos three π‘₯ equals sin π‘₯ over four is therefore π‘₯ is equal to two πœ‹ over 13 plus eight π‘›πœ‹ over 13, and π‘₯ is equal to negative two πœ‹ over 11 plus eight π‘›πœ‹ over 11, where 𝑛 is an integer.

Now you may recall that there were two possible avenues we could have followed from our trigonometric identity for sin 𝐴 minus sin 𝐡. We could also have tried to solve sin 𝐴 minus 𝐡 over two equals zero. But in fact, this gives us exactly the same solution set as we found solving the cos of 𝐴 plus 𝐡 over two is equal to zero, so our solution is complete.

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