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.

5 cos² x - 3 = 0

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 two pi employing exact values whenever feasible or providing approximations accurate to four significant figures. Our equation is seven minus cosine squared X equals zero answer. Our answer choices are answer choice A X equals the solution set 0.57962 point 5623.721 and 5.704. Answer choice B X equals the solution set. Pi divided by 12 5 pi divided by 12 3 pi divided by 4, 13 pi divided by 12 17 pi divided by and seven pi divided by four answer choice C X equals the solution set 0.99861 point 3422.6993 point 186 and 4.95. Answer choice D is no solution. All right. So what we want to do with this, we've only got one cosine function in here and it's cosine squared X but we still want to start off by getting that isolated on one side of the equal sign. So we'll subtract seven from both sides of the equation. And then we'll have a negative 10 cosine squared X equals negative seven. Then we will divide both sides by negative 10. And we'll have on the left cosine squared X equals negative seven divided by negative 10 is a positive 7/10. And then to get that cosine squared X down to a regular cosine of X, we take the square rep and what we do to one side we do to the other. So now we will have the cosine of X equals plus or minus the square rut of 7/10. So now we have it solved for the cosine of X and that's not any exact value. So this one, we will be doing the approximations accurate to four significant figures. Now to solve for X, we recall from previous lessons that we can use the inverse cosine function. So we actually have two different things. We're setting it equal to and we say that X equals the inverse cosine of the positive square of 7/10. And then we will be doing X equals the inverse cosine of negative square rut of 7/10. So we can plug that into the calculator. We'll start with the inverse cosine of the square root of 7/10 which is positive, we plug that in and we are going to get 0.57963. And we think about that for a moment, if we draw a rough sketch of a coordinate plane, we have a vertical y axis, a horizontal X axis, they come together at the origin in the middle. Now, we're talking about everything being in radiance here. And if you didn't get that answer, be sure to check your mode if you're in degrees or radiance and talking about radiance, we know that the left hand side of the X axis, the part going toward negative infinity, that's pi which is approximately 3. and about half of 3.14 is about 1.5. And that's going to be the positive part of the Y axis. And so if we have an angle of 0.57963, that's going to be in the first quadrant. So we have a quadrant one solution of 0.57963 for X. However, cosine is also positive in quadrant four. So because we have our quadrant one angle, that can be our reference angle for figuring out what it is in quadrant four. And so we'll take two pi minus our reference angle and to be very specific, that's the inverse cosine of the square rut of 7/10. And we plug that into a calculator and we are going to get 5.7035, that's going to be our quadrant four solution here for our cosign. And now we have another answer where we have X equals the inverse cosine of negative square at 7/10. We plug that into the calculator and we get 2.5619. And that is between 1.5 and 3.14. We can see that that is our quadrant two solution which makes sense because we're taking a negative, we, for our cosine, we have a negative there. And so that makes sense, our quadrant two solution. Now we just need to figure out our quadrant three solution. So in order to do that, we can use our reference angle from quadrant one again, and we can take that we can take pi plus our inverse cosine of the square rut of 7/10. And that is going to equal 3.7212, there's our quadrant three solution. So we have four solutions here. And now it's just a matter of writing these to four significant figures. So our quadrant one smallest angle here, our smallest X solution is zero point 579 and now six is the fourth significant figure. It looks to the three and that's just going to say a six, the next one, our quadrant two angle we have 2.56. Now the one is our fourth significant figure. The nine is going to have that round up. So it's a two, our quadrant three angle 3.72. That uh one is our fourth significant figure. It looks to the two and it remains a one. And then we have our quadrant four angle 5.70. Now the three is the fourth significant figure. It looks to the five and it rounds up to a four. And that is our solution set for X. We look at our answer choices and this matches with answer choice. A well done. We'll catch you on the next one.

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. 5 cos² x - 3 = 0

Key Concepts

Trigonometric Identities

Solving Quadratic Equations

Interval Notation

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