In Exercises 54–67, solve each equation on the interval [0, 2𝝅)....

Question

In Exercises 54–67, solve each equation on the interval [0, 2𝝅). Use exact values where possible or give approximate solutions correct to four decimal places.
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sin 2x = √ 3 sin x

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 2 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 the square rut of multiplied by the cosine of B equals eight. Sign of to be our answer choices are answer choice A B equals the solution set zero pi divided by four pi and three pi divided by four answer choice B B equals the solution set. Pi divided by six pi divided by 25 pi divided by six and three pi divided by two answer choice C B equals the solution set pi divided by three pi divided by 22 pi divided by three and three pi divided by two and answer choice D no solution. All right. So we've got a cosine cosine on one side equal to a sign on the other side. And we see that we have the square rat of 192 in front of our cosine of B on the left. Well, we know that the coat that excuse me, the square rut of 192 is equal to the square rut of 64 multiplied by three all under the radical. And we can take the square of 64 that's eight. So this simplifies to eight square at three. So in front of our cosine B, we have eight square at three, cosine B equals eight sine two B. Now let's take a look at the right hand side. We see that we're taking the sign of two B. And we recall from previous lessons, the double angle formula where the sign of two multiplied by our angle, which is B here is equal to two multiplied by the sign of that angle. So the sign of B multiplied by the cosine of B. And so we can rewrite it that way. Let's bring down the left hand side eight square at three cosine B equals. And then on the right, we have eight multiplied by the quantity of now instead of sine of two B and we have two sign B cosine B and we can divide everything here by eight, both sides. And that's going to get those eights to cancel out. So this cleans up nicely on the left, we've got the square of three cosine B equals on the right two sine B cosine B. Now it's time to bring these to the same side of the equal sign, we can subtract two sine B cosine B from both sides of the equation. And then on the left, we'll have the square it of three cosine B minus two sine B cosine B equals on the right, just zero. We can factor out this common factor, the cosine B. And so we'd have cosine B multiplied by. And then when we factor that out, what's left is square at three minus two sign B and that all equals zero. Well, when we're multiplying two factors and they're equal to zero, we can set each of the two factors equal to zero and solve. So here we've got a cosine B equals zero and we've got square at three minus two sign B equals zero. So let's start with the cosine of B equals zero. Where is it true that the cosine of an angle is equal to zero? Well, we recall from previous lessons and from tables in the book that this is true at the quadrant angles of pi divided by two and the three pi divided by two. So those are two solutions for our angle B pi divided by two and three pi divided by two. Now let's take a look at the second one, the square of three minus two sine B equals zero. We can do a little rearranging here. We can subtract the square of three from both sides and we'll have negative two sine B equals negative square at three. We can divide both sides by negative two and will have the sign of B equals. And on the right, it'll be a positive square at three divided by two. All right. So where is it true that the sign of an angle equals the square out of three divided by two? And we recall from those same previous lessons and from the tables in the book that this is true when B equals pi divided by three and pi divided by three is in the first quadrant, where else is sine positive? Well, it's also positive in the second quadrant. So we can take to find for our quadrant two solution, we can take pi minus our reference angle from angle one which is pi divided by three. And that will give us two pi divided by three, which is our last solution here. So we've got four angles. We have pi divided by three pi divided by two. We have two pi divided by three and three pi divided by two. We look at our answer choices and this matches with answer choice. See, well done. We'll catch you on the next one.

In Exercises 54–67, solve each equation on the interval [0, 2𝝅). Use exact values where possible or give approximate solutions correct to four decimal places. __ sin 2x = √ 3 sin x

Key Concepts

Trigonometric Identities

Solving Trigonometric Equations

Interval Notation

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