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.

tan x sec x = 2 tan x

Accepted Answer

Welcome back. I am so glad you're here. We are 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 acc accurate to four significant figures. Our equation is two square at three tangent X equals three C can't X tangent X. Our answer choices are answer choice A X equals the solution set zero pi divided by four pi and pi or excuse me, seven pi divided by four answer choice B is X equals the solution set zero pi divided by three pi and five pi divided by three answer choice C X equals the solution set zero pi divided by six pi and 11 pi divided by six. And answer choice D is no solution. All right. Looking at our equation here, let's start off by bringing everything to the same side of the equal sign. So we'll take from the right hand side, our three C can X tangent X will subtract that and we'll subtract that from both sides. So on the left, it will read two square at three tangent X minus three C can't X tangent X equals zero. And now we have a common factor in between these two terms being subtracted on the left. They both have a tangent X. And so we can factor that out in front. So we'll have our tangent XX multiplied by the quantity of what's left when we factor it out is two square at three minus three C X. And when we have two factors multiplied by each other equal to zero, we can set each of those two factors equal to zero. So for one, we have the tangent of X equal to zero. And for the other, we have two square at three minus three C can't X equal to zero. And we can simplify the one on the right a little bit more. We can subtract two square at three from both sides of the equation. On the left, we have negative three C can't X and that equals on the right negative two square at three. Now we can divide both sides by negative three. And on the left, we have the ent of X equals and on the right negative two square at three divided by negative three is a positive two square at three divided by three. And we recall from previous lessons that the C can of X is equal to one divided by the cosine of X. And we can substitute in for the C can of X, our one divided by the cosine of X, I'm going to rate that up here where there's a little more room. So we have one divided by the cosine of X equals the two square at three, divided by three. And we know that when we have two fractions that are equal to, to each other, the reciprocals of those two fractions are equal to each other. So on the left, we have the cosine of X equals on the right, we have three divided by two square at three. But we want to rationalize that we don't want that radical in the denominator. So we multiply both the numerator and the denominator by the square of three. And so in the numerator, we have three square at three and that's divided by two, multiplied by the square out of three, multiplied by the square out of three is two multiplied by three, which is six. And so this is going to simplify to be the cosine of X equals the square of three divided by two. So now we have two equations, the tangent of X equals zero and the cosine of X equals the square of three divided by two. And now we just need to recall previous lessons and the tables in the book where is tangent of an angle equal to zero. Well, that's going to be at pi and at zero and there's two solutions for the tangent of X equals zero. And from those same tables in previous lessons, where is the cosine of X equal to the square root of three divided by two. Well, that's when X is equal two pi divided by six. And that's in the first quadrant, where else is the cosine positive? And it's also positive in the fourth quadrant. So if we take pi or excuse me, two pi minus our reference angle from quadrant one. So minus pi divided by six, that's going to give us 11 pi divided by six for our quadrant four solution. So there are our four solutions for X. We look at our answer choices and this matches with answer choice. C 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. tan x sec x = 2 tan x