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. 2 sin² x = 3 - sin x

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 pi inclusive of zero, but not inclusive of two pi employing exact values whenever feasible or providing approximations accurate to four significant figures. Our equation is five sine squared X equals 12 sine X minus seven. Our answer choices are answer choice A X equals the solution set seven pi divided by six and 11 pi divided by six. Answer choice B X equals three pi divided by four. Answer choice C X equals pi divided by two and answer choice D no solution. All right. So to solve this, let's first start out getting everything together on the same side. So we can subtract 12 sine X minus seven from the right hand side and subtracting that whole thing from the left hand side as well. And so on the left, we'll have five sine squared X and then distributing that subtraction, it will be minus 12 sine X and plus seven all equal to zero on the right hand side. Now, we can see that we have a nice quadratic. And to make this a little bit more intuitive to solve, we can set the sign of X equal to any variable. I'm going to choose U, then we'd have five, not multiplied by the sine squared of X, but multiplied by U squared minus 12, not multiplied by the sign of X, but multiplied by U plus seven equals zero. Now we can factor this going to do the reverse of the foil process if you need any refreshers on factoring, definitely check out previous lesson videos. I'm going to factor this very quickly. So in the first set of parentheses, that's going to factor to be five U minus seven and that's multiplied by the quantity of U minus one. Now we have two factors multiplied that are equal to zero. So we can set each factor equal to zero to solve for you. We have five U minus seven equals zero and U minus one equals zero for five, U minus seven equals zero. We'll add seven to both sides, then we have five U equals seven. Then we divide both sides by five and we get U equals 7/5. Now for the second one, U minus one equals zero, we add one to both sides and we get U equals one. But recall, we're not trying to solve for you. We are trying to solve for X and we know that U is equal to the sign of X. So we rewrite this as the sign of X equals a positive 7/5 and the sign of X equals one. Now, it's just a matter of solving for X. Now, for the first one, the sign of X equals a positive 7/5. Well, we could use our calculators, we, we know that X equals the inverse sign of whatever is on the right hand side. But when we put the inverse sign of 7/5 into our calculator, we are going to get a domain error. And that is because the domain of our inverse sign is from negative 1 to 1 and 7/5 that is 1.4, that is more than one. So it's outside of our domain. Now, we can take a look at the sign of X equals one, not one, we can use previous lessons. And at the tables in the textbook to see where the sign of X is equal to one. And that is at the quadrant angle of pi divided by two. And that's the only location where the sign of X is equal to one. So that's our solution. We look at our answer choices and this matches with answer choice. C well done. We'll catch you on 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. 2 sin² x = 3 - sin x

Key Concepts

Solving Quadratic Trigonometric Equations

Unit Circle and Interval Restrictions

Exact Values and Approximate Solutions

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