Exercises 39–52 involve trigonometric equations quadratic in f...

Question

Exercises 39–52 involve trigonometric equations quadratic in form. Solve each equation on the interval [0, 2𝝅). 2 sin² x = sin x + 3

Accepted Answer

Everyone. In this problem, we have the trigonometric equation. Four sine squared of B is equal to sine B plus five. And we're asked to use algebraic techniques to solve this in the interval from 0 to pi excluding two pi. And to give the answer in radiance, we have four answer choices. Option A is that B is equal to pi divided by two. Option B B is equal to three pi divided by two. Option C B is equal to pi and option DB is equal to two pi. So let's get started by writing up this equation again or Stein squared B is equal to sine B plus by now, this looks a little bit scary, but we have dealt with equations very similar to this. OK. So we have a squared term, we have a linear term and we have a constant term. So this is just like a quadratic equation where we have X squared plus X plus one equals zero, something like that. How do we solve those? We know how to solve those with the quadratic formula? So let's first move everything to the left hand side and have just zero on the other side. OK. So we have that four time squared B minus sign B minus five is equal to zero. And to make this simpler, what we're gonna do is do a little substitution. We're gonna substitute and let X equal sign of B. If we substitute this in our equation, that equation is gonna look just like a quadratic that we know how to solve. We're gonna be able to get that X value and then come back and relate it to sign of B. OK. So substituting this in, we get that four X squared minus X minus five is equal to zero. Let's go ahead and use the quadratic formula here. OK? And recall that we use the quadratic formula. We need our A B and C terms. OK. So the A term is gonna come from the coefficient corresponding with the X squared term. The B term is gonna be the coefficient corresponding with the X term and the C term is gonna be that constant term at the end. And so X is gonna be equal to negative B pull off or minus the square root of B squared minus four AC all divided by two A substituting in our values B is going to be negative one. So negative B is gonna be positive one plus or minus the square root of B squared, negative one squared minus four multiplied by A which is four multiplied by C which is negative five, all divided by two, multiplied by A which is four. If we simplify, we have one squared minus four, multiplied by four, multiplied by negative five, this gives us one plus 80. And so we get 81 under our square root. So we have X is equal to one plus or minus the square root of 81 divided by eight. We know that the square root of 81 is equal to nine. And so we get that this is equal to one plus or minus nine divided by eight. OK? And so we're gonna get two X do you here and we either get X is equal to one plus nine divided by eight. OK? Which is gonna be 18th or dividing numerator and denominator by two, we can simplify to five quarters or we get that X is equal to one minus nine divided by eight and that's negative eight divided by eight which gives us negative one. OK? So we have these two possible X values but remember we're not solving for X and we're solving for B. And so what we wanna do is use these values for X, translate them back into our sine B equation. So we know that sin of B or X is equal to sin of B. Sorry. So what we have is that sine of B is equal to five quarters from the first X value or we have that sign of B is equal to negative one. From the second X value starting on the left hand side with this first equation sine of B is equal to five quarters. Well, there's actually no solution here. OK. Remember that Sine of B has a maximum of one and we can write that negative one is less than, or equal to sign data which is less than or equal to one. OK? It's never greater than one. And so there is no solution, There is no B value that will give us this value. OK? So no solution from that half, moving to the second half, Sine B is equal to negative one. OK? And we're between zero and two pi there's a couple of ways you could think about this. OK? You could remember the values where Sine B is equal to negative one. Some people like to use a unit circle and that's a great option as well. What I'm gonna do is I'm just gonna draw a rough sketch of Sine. OK? If you recall what the Sine graph looks like, we know it starts at 00, it goes upwards, back downwards and, and completes its cycle. Now it crosses a pi and then at two pie. And so there's minimum where we have Sine B equal to negative one is gonna be at three pi divided by two. OK? So if you can remember the shape of your sign and your cosine curves and the key points on them, that can be a really effective way to solve some of these equations as well. And so we get here that B is gonna be equal to three pi divided by two. And that's the only solution on this interval that we were given. If we go up to our answer choices and compare what we found. We see that the correct answer is option B. Thanks everyone for watching. I hope this video helped see you in the next one.

Exercises 39–52 involve trigonometric equations quadratic in form. Solve each equation on the interval [0, 2𝝅). 2 sin² x = sin x + 3

Key Concepts

Quadratic Form in Trigonometric Equations

Solving Trigonometric Equations on a Specific Interval

Using the Unit Circle to Find Angle Solutions

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