In Exercises 97–116, use the most appropriate method to solve eac...

Question

In Exercises 97–116, use the most appropriate method to solve each equation on the interval [0, 2𝝅). Use exact values where possible or give approximate solutions correct to four decimal places.

sin 2x + sin x = 0

Accepted Answer

Welcome back. I am so glad you're here. We're asked to apply the most suitable technique to determine the solution of the following equation within the interval from 0 to pi inclusive of zero but not of two pi employing exact values whenever feasible or providing approximations accurate to four significant figures. Our equation is three sine two X plus three square at three cosine X equals zero. Our answer choices are answer choice A X equals the solution set pi divided by 23 pi divided by 24 pi divided by three and five pi divided by three answer choice B X equals the solution set. Pi divided by 23 pi divided by 27 pi divided by six and 11 pi divided by six. Answer choice C X equals the solution set. Pi divided by 23 pi divided by 25 pi divided by four and seven pi divided by four and answer choice D X equals the solution set pi divided by 23 pi divided by four and three pi divided by two. All right. So looking at our equation here, we recognize in here we've got three, multiplied by the sign of two X. And we recall from previous lessons, our double angle formula which says that the sign of two multiplied by an angle. So in this case, X is equal to two multiplied by the sign of that angle. So here, the sign of X multiplied by the cosine of X. And so we can use substitution for that sign of two X. We'll still have that three, but now it's multiplied by the quantity of and now plugging in two sine X cosine X. And then that is all plus three square at three, cosine X still all equal to zero. Now, we can distribute that three. So three multiplied by two is going to be six. So we have six sine X cosine X plus three square at three cosine X equals zero. And we have a common factor here. The common factor is cosine X. So we can factor that out in front. We have the cosine of X multiplied by the quantity of what's left. Once we've factored out the cosine of X, we still have six sine X plus three square at three, all equal to zero. Now, when we have two factors multiplied by each other that are equal to zero, we can set each of the two factors to zero and solve. So the first factor, the cosine of X is equal to zero. And the second one, we have six sine X plus three square at three equals zero. We can rearrange that one a little bit, we can subtract three square at three from both sides of the equation. And then we'll have six sine X equals negative three square at three. And now we can divide both sides of the equation by six. And on the left, we just have the sign of X and that's equal to, on the right negative three square at three divided by six, simplifies two negative square at three divided by two. So we have two equations, the cosine of X equals zero and the sign of X equals negative square at three divided by two. So now we want to recall previous lessons and recall the tables in the books that tell us and we'll start with the cosine of X equals zero. The cosine of X equals zero at pi divided by two and at three pi divided by two. So there are two solutions. And now where does the sign of X equal negative square at three divided by two? Well, if we're talking about the sign of X equaling the positive square at three divided by two, that is when X equals pi divided by three. But here we're talking about the sin of X equaling a negative square square root three divided by two. So we can use pi divided by three from quadrant one as our reference angle. And now where is sine, negative sine is going to be negative in quadrants three and four. So using our reference angle we want to take for the quadrant three angle, we'll take pi plus our reference angle pi divided by three. And that's going to give us four pi divided by three. There is our quadrant three solution for the sign of X equals negative square at three divided by two. That's when XX equals four pi divided by three. Now, for quadrant four, we'll take two pi minus our reference angle pi divided by three. And that's going to give us five pi divided by three and those are our four solutions. Here, we look at our answer choices. Four pi divided by 23, pi divided by 24 pi divided by three and five pi divided by three. And that matches with answer choice a well done. We'll catch you on the next one.

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In Exercises 97–116, use the most appropriate method to solve each equation on the interval [0, 2𝝅). Use exact values where possible or give approximate solutions correct to four decimal places. sin 2x + sin x = 0