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.

tan x = 2 cos x tan 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 six cosine X tangent X is equal to three square at three tangent. Our answer choices are answer choice. A no solution answer choice B X equals the solution set zero pi divided by three pi divided by two and five pi divided by three answer choice C X equals the solution set 0.54550 point 6566 and 0.7878. And answer choice D X equals the solution set zero pi divided by six pi and 11 pi divided by six. All right. So we've got cosines and tangents and one tangent on each side of the equal sign. Let's start off by bringing the tangents on the right to the left hand side. So we'll subtract three square at three tangent X from both sides of the equation. And there are a lot of ways to go about doing this. So it's OK if you may have started somewhere else. So then on the left, we'll have six cosine X tangent X minus three square at three tangent X equals on the right zero. And now we have a common factor between these two terms. We have a tangent X in both of them. And we can factor that out. So we have tangent multiplied by the quantity of, and what's left is six cosine X minus three square at three, that's all equal to zero. Now, when we have two factors being multiplied together that equal zero, we can set both of those two factors equal to zero. So we have the tangent of X equals zero and we have six cosine X minus three square at three equals zero. We can do a little rearranging with the second one. We can add three square at three to both sides. And then on the left, we'll have six cosine X equals. And on the right, we have three square at three and we can divide both sides by six. And we have that the cosine of X equals the right hand side simplifies two square at three divided by two. So we've got these two equations here and these are going to be exact values because we recall from previous lessons in the tables in our books that the tangent of X is equal to zero when our angle X is at zero. And at pi and we recall from those same lessons and same tables that the cosine of X is equal to the square root of three divided by two when X is equal to pi divided by six. Now that is in quadrant one, where else is cosine positive, it's also positive in quadrant four. And so if we use pi divided by six as our reference angle from quadrant one, we find our quadrant four angle by taking two pi minus the reference angle pi divided by six. And that's going to get us 11 pi divided by six. And those are our four angles here. So we have zero followed by pi divided by six, followed by pi followed by 11 pi divided by six. We look at our answer choices and this matches with answer choice. D well done. We'll catch you on the next one.

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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. tan x = 2 cos x tan x