In Exercises 53–62, solve each equation on the interval [0, 2𝝅)....

Question

In Exercises 53–62, solve each equation on the interval [0, 2𝝅).
                   
sin x + 2 sin x cos x = 0

Accepted Answer

Hello, everyone. We are asked for the following trigonometric equation to solve for the interval to 2 pi and give the answer an exact radiant measure. The equation is three cosine of XX plus six sin of X cosine of X equals zero. Our answer choices are A X equals the set three pi divided by 45 pi divided by 47 pi divided by four BX equals the set pi divided by 24 pi divided by 33 pi divided by 25 pi divided by three CX equals the set pi divided by 27 pi divided by 63 pi divided by 2, 11 pi divided by six D no solution. Looking at the original equation, we have three cosine X plus six sine of X cosine of X equal to zero. I notice that I have a greatest common factor and that three cosine of X could be factored out of each term. So I'm gonna do so and I will get three cosine of X multiplied by the parentheses. One plus two sine of X closed parentheses equals zero. From here, I can set each factor equal to zero. So I have three cosine of X equals zero and one plus two sin of X equals zero. And we're gonna find the values for X that make these things true. So starting with three cosine of X equals zero, if I divide both sides by three, I get the cosine of X equals zero. And if I look at a unit circle to find where is the cosine zero? What angles I find that it is zero at pi divided by two and three pi divided by two. So I'm boxing those in. For now, we're gonna verify that they work in our equation. But for now those look like two of our angles. Next for one plus the two sign of X equals zero, subtract one from both sides. Two sine X equals negative one divide by two. So we need to find angles when the sign of X is negative one half, going back to my unit circle, I find that sign of seven pi divided by six is negative one half and the sign of 11 pi divided by six is negative one half. So we now have four angles to try. So we're gonna do that one at a time starting with pi divided by two. I'm gonna plug this back into my original equation. So I have three multiplied by the cosine of pi divided by two plus six, multiplied by the sign of pi divided by two, multiplied by the cosine of pi divided by two And if this is a true solution, it will equal zero. So substituting in our values, we have three multiplied by, we know the cosine of pi divided by two is zero plus six, multiplied by the sign of pi divided by two, which is one multiplied by the cosine of pi divided by two, which is zero has to equal zero. Well, zero, multiplied by anything is zero. So zero plus zero equals zero. And we know that pi divided by two, it works. Similar thing is going to happen with three pi divided by two because the cosine is zero. But we're gonna plug it in anyway. So three multiplied by the cosine of three pi divided by two plus six, multiplied by the sign of three pi divided by two, multiplied by the cosine of three pi divided by two has to equal zero. Again, knowing that three pi divided by two, the cosine of it is zero. I can jump to saying zero plus zero equals zero. So those two angles check, we know that pi divided by two and three pi divided by two are solutions that work in our equation testing seven pi divided by six. These ones will take a little bit more work because they don't equal to zero. But it'll be all right. So we will start with the code three multiplied by the cosine of seven pi divided by six plus six, multiplied by the sign of seven pi divided by six, multiplied by the cosine of seven pi divided by six and that must equal zero. So three and then the cosine of seven pi divided by six is negative radical 3/2 or divided by two plus six, multiplied by the sign of seven pi divided by six, which is negative one half multiplied by the cosine of seven pi divided by six, which we said was negative radical three divided by two. And we want this to equal zero. So using my math rules and multiplying my fractions, I have negative three radical three divided by two plus, I'm gonna cross reduce my two and my six and then a negative multiplied by a negative is a positive. So we will have a positive three radical three divided by two and negative three radical three divided by two plus a positive three radical three divided by two will equal zero. So seven pi divided by six also works last 1. 11 pi divided by six. So we'll have three multiplied by the cosine of 11 pi divided by six plus six multiplied by the sign of 11 pi divided by six, multiplied by the cosine of 11 pi divided by six. And this must equal zero for it to be true. Going back to our values three multiplied by what is the cosine of pi divided by six. Well, it appears to be radical three divided by two and then adding six multiplied by the sign of 11 pi divided by six which is negative one half multiplied by the cosine of 11 pi divided by six which was positive radical three divided by two and this needs equal zero. So doing our multiplication, we have three radical three divided by two plus a negative. And then I'm gonna cross reduce three radical three divided by two, three radical three divided by two. There we go. And when we add those together, this will equal zero. So our answers are pi divided by 23 pi divided by 27 pi divided by 11 pi divided by six. And that matches answer choice C have a nice day.

In Exercises 53–62, solve each equation on the interval [0, 2𝝅). sin x + 2 sin x cos x = 0

Key Concepts

Trigonometric Identities

Factoring Trigonometric Equations

Interval Notation

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