In Exercises 97–116, use the most appropriate method to solve ...

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. 5 cot² x - 15 = 0

Accepted Answer

Hey, everyone in this problem, 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 excluding two pi, OK. Giving exact values whenever possible or providing approximations accurate to four significant digits. And the equation we're given is 69 minus cotangent squared of X is equal to zero. We have four answer choices. Option A has that X is a set containing four values. Option B is the same as option A, it just differs in what those four values are. Option C is a set containing two possible values for X and option D is that there is no solution. So let's start by rewriting what we were given. We have 69 minus 23 contingent squared of X is equal to zero. Now, we don't want to get overwhelmed by the fact that we have this trigonometric function in here. OK. We want to solve this like any other equation. We wanna isolate X by getting all of the numbers on one side and having X left on the other. So let's start by moving this 23 cotangent squared X to the other side, then we can write 23 co change and squared of X is equal to 69. We're gonna divide both sides by 23. You get that coin and squared of X is equal to three. Now, we need to take the square root. OK. So we have a contingent of X is equal to. And what's really important to remember is that when you take the square root, you get both the positive and the negative route. OK? And this is going to change how many solutions we have. OK. So in this case, we aren't told any more information, OK? To figure out whether cotangent could be positive or negative. So we're gonna assume that it can be both, we have both the positive and the negative route and we're gonna have to carry it forward with those. Now, let's recall, OK. The tingent of X is equal to one divided by tangent of X. So what this tells us is that if we have co tangent of X is equal to plus or minus the square root of three. So we can write that the tangent of X must be equal to plus or minus one divided by the square root of three. All right. So, well, let's start by finding our reference angle and our reference angle is gonna be between zero and pi divided by two. It's gonna be the angle that solves this, that is in that range. And let's think about our special triangles. OK. Remember that we can draw a special triangle. And for this triangle, we wanna have side lengths of one in the square of three. OK. Well, we have a triangle that's like if we have sides one in the square root of three, then the hypotenuse will be two. And recall that when we have a triangle like this, the two angles will be pi divided by three and pi divided by six. And we know that pi divided by six will be across from side length one because that's the smallest angle and the smallest side and the smallest angle is always across from the smallest side. So we have our special triangle here and we want to find the angle that when we use tangent gives us one divided by the square root of three. And a recall tangent is going to be the opposite side divided by the adjacent side. And so the angle we're looking for is pi divided by six. So our reference angle is going to be pi divided by six. What other solutions do we have? OK. Do we have more solutions? And what are they? And that's really, really important. So let's go down. So we have a tangent could be positive or negative if tangent is positive or negative, that means that we can be in any quadrant. So if we draw our plane here, I recall that we have quadrant one in the top right and then moving um counterclockwise quadrant two quadrant three in quadrant four, a reference angle is in quadrant one always. And this is pi divided by six. Now, for all of our other angles, OK. And all of the other quadrants, the reference angle is going to be measured from the closest X axis into that quadrant. So in quadrate two, we're gonna go from the negative X axis upwards. OK? Pi divided by six units in quadrant three, we go from the negative X axis downwards pi divided by six in a quadrant before we go from the positive X axis downwards I divided by six gradients. So we have these four possible solutions. But when we actually write out our solution, we want to remember that the angles are always measured from the positive X axis. So the first solution from quadrant one, X one is gonna be pi divided by six. Now the second. OK. Well, we're gonna go from the positive X axis counter clock wax until we hit that angle. This is gonna be X two. And we can see that X two. Well, if we go all the way to the negative X axis that would be pi radiance. But we're stopping pi divided by six radiant short. So X two is gonna be pi minus pi divided by six which gives us five pi divided by six, moving to quadrant three. Again, we're starting at the positive X axis. We're going counterclockwise until we hit that angle. Hey, this is gonna be X three. And for X three, we're going pi units scale all the way to the negative X axis making the straight line and then we're going pi divided by six more. So X three is gonna be pi plus pi divided by six, which gives us seven pi divided by six. And finally, for our solution in quadrant four again from the positive X axis counterclockwise until we hit that angle. We can see that if we went all the way around to form one complete circle, that would be two pi radiant. But we're stopping pi divided by six radiant short. And so X four is going to be two pi minus pi divided by six, which gives us 11 pi divided by six. OK. So if you're ever confused about whether you should be adding your reference angle, subtracting your reference angle, whether you should be doing it from pi or two pi drawing it out, I find it is a really helpful way to visualize what's happening and figure out whether you should be adding or subtracting and from pi or two pi. All right. So we found these four solutions X one, X two, X three and X four all to that original equation. If we go back up to our answer choices, OK, we know it's gonna be option A or B because they each have four options and the correct answer here is going to be a because those values match with what we found we have. That X is gonna be the set containing pi divided by 65 pi divided by 67 pi divided by six and 11 pi divided by six. Thanks everyone for watching. I hope this video helped see you in the next one.

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. 5 cot² x - 15 = 0

Key Concepts

Trigonometric Equations

Cotangent Function and Its Properties

Solving Quadratic Equations in Trigonometric Form

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