In Exercises 12–18, solve each equation on the interval [0, 2𝝅)....

Question

In Exercises 12–18, solve each equation on the interval [0, 2𝝅). Use exact values where possible or give approximate solutions correct to four decimal places.2 sin² x + cos x = 1

Accepted Answer

Hello, everyone. We are asked to apply the most suitable technique to determine the solution for the following equation within the interval 0 to 2 pi including zero but not including two pi employing exact values whenever feasible or providing approximations accurate to four significant figures. The equation we are given is two multiplied by the sine squared of X plus open parentheses, two minus radical, two closed parentheses multiplied by the cosine of X equals two minus radical two. We are given four answer choices with sets for X values. Each answer choice has X equal to zero. But the other two values vary to begin with. Recall that the sign squared of X equals one minus the cosine squared of X. We are going to use this to substitute it into our equation so that we're working with just one function. So we're just gonna work with cosines. So this would give us two multiplied by in parentheses, one minus cosine squared X closed parentheses plus open parentheses, two minus radical, two closed parentheses, cosine XX equals two minus radical two from there. I'm going to distribute the two into the parentheses. So we will have two minus two cosine squared X plus open parentheses, two minus radical two closed parentheses, cosine X equals two minus radical two. From here, we're gonna try to get the right hand side equal to zero. So I'm gonna start by subtracting two from both sides which will leave us with negative two cosine squared X plus open parentheses two minus radical two closed parentheses, cosine X equals negative radical two. I'm going to add radical two to both sides. And I have negative two cosine squared X plus open parentheses, two minus radical two closed parentheses, cosine X plus radical two equals zero. And I don't want my leading coefficient to be negative. So I'm going to divide everything by negative one and that will give us two cosine squared X minus open parentheses, two minus radical two closed parentheses, cosine X minus radical two equals zero. From here, we need to find out what could this equals. We need to find what our XS could be. We're gonna use the quadratic formula in order to do so we need to first do some substituting. So we are gonna say that Y equals the cosine of X. And when we do, we can rewrite this equation as two Y squared minus the parentheses, two minus radical, two closed parentheses, Y minus radical two equals zero. So using the quadratic formula, my A value is two, my B value is negative parentheses, two minus radical, two close parentheses. And my C value is negative radical two. So Y equals the opposite of B. So that'd be two minus radical two plus or minus the square root of B squared. So in brackets, I'm gonna put negative open parentheses, two minus radical two closed parentheses, close brackets squared minus four, multiplied by A which is two multiplied by C which is negative radical. Two closed parentheses divided by two A. So two multiplied by two. So let's see what we can simplify. So Y equals two minus radical two plus or minus the square root and underneath. So if I square the brackets which have negative open parentheses, two minus radical two, since it's a negative being squared, it'll come out to a positive. So that part I won't worry about. And then I recall when I square a binomial I can square the first term. So two squared is four square, the last term. So negative radical two squared would be a positive two. And then I take the product of the two terms together, which would be negative two radical two and double it. So that would be negative for radical two. And then I have minus four multiplied by two, multiplied by negative radical two. So negative four multiplied by negative radical two will give us a positive four radical two multiplied by two would be plus eight radical two, all of that divided by two multiplied by two, which is four. All right. Let's keep going. Let's see what else we can combine in there. So Y equals two minus radical two plus or minus the square root of. Let's see, we're gonna combine some like terms four plus two is six, negative four radical two plus eight radical two is a positive four radical two and it will be divided by four. So now I need to figure out how am I going to get the things that are under that radical out? I'm gonna start by splitting up this binomial. So I'm gonna do that to the left here in another little column. So I have six plus four radical two that I want to break apart and I can break this apart to four plus two radical two plus two radical two plus two. So I broke the six into four and two and four radical two into two radical two plus two radical two from there. If I think of the number two as the product of radical two multiplied by itself, I can rewrite this expression as four plus two radical two plus two radical two plus in parentheses, radical two multiplied by radical two, close parentheses. So now I'm gonna factor something out of are terms by grouping. So with my first two terms, so four plus two radical two factor out of two. So that'll give me two multiplied by two plus radical two in parentheses. And then if I group my second set of terms, I can factor a radical two out of them. So I have plus radical two multiplied by the parentheses, two plus radical two closed parentheses, rewriting this, I get two plus radical two multiplied by two plus radical two. So this can be two plus radical two in parentheses squared. So we're gonna substitute that in under our radical symbol in our quadratic formula. So we get Y equals two minus radical two plus or minus the square root of two plus radical two in parentheses squared, divided by four. From here, I know that the square root of a square will cancel it out. So we will have Y equals two minus radical two plus or minus in parentheses, two plus radical two closed parentheses divided by four. At this point, I need to split this into two equations. So I'm gonna follow that up to here. So I'll have Y equals two minus radical two plus two plus radical two divided by four. And or I will have Y equals two minus radical two minus the parentheses, two plus radical two, close parentheses divided by four. So looking at the 1st 12 minus radical, two plus two plus radical two divided by four, the radical twos will cancel out and this would be two plus two, which is four divided by four, which is one. So this means one of our Y values is one on the other. I'm gonna distribute the negative into the parentheses that follows. So I will have that Y equals two minus radical two minus two minus radical two divided by four, combining like terms the two and the negative two will cancel out. So I have negative two radical two divided by four. And simplifying this gives me negative radical two divided by two. So now that we have values, we're gonna go back to Y equals the cosine of XX. So if Y is one, one equals the cosine of X, well, when the cosine is one, it's either at zero or two pi since our interval says that two pi is not a solution in the solution set, that means X has to equal one for this. Now when Y equals negative radical two divided by two. So when that is the cosine of X checking a unit circle, we find that that occurs at the value where X is three pi divided by four and at five pi divided by four. So these are our X values. This is the solution. So X equals zero three pi divided by pi divided by four. And that is answer choice. A have a nice day.

In Exercises 12–18, solve each equation on the interval [0, 2𝝅). Use exact values where possible or give approximate solutions correct to four decimal places.2 sin² x + cos x = 1

Key Concepts

Trigonometric Identities

Solving Trigonometric Equations

Interval Notation

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