Solve each equation (x in radians and θ in degrees) for all ex...

Question

Solve each equation (x in radians and θ in degrees) for all exact solutions where appropriate. Round approximate answers in radians to four decimal places and approximate answers in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures.
2 cos² x + cos x ― 1 = 0

2 cos² x + cos x ― 1 = 0

Accepted Answer

Hey, everyone in this problem, we're asked to find the solutions of the given equation and express our answer in radiance using the least possible nonne angle measure. The equation we're given is two sine squared of X plus sine X is equal to one. We have four answer choices. Option A, we have two solutions. Five pi divided by six plus two N pi and three pi divided by two plus two N pi where N is an integer. Option B again two solutions pi divided by three plus two N pi and three pi divided by two plus two N pi again N is an integer. Option C pi divided by six plus two N pi five pi divided by six plus two N pi and three pi divided by two plus two N pi where N is an integer. And finally, option D is that there is no solution. So let's start, we're gonna rewrite the equation. We were given, we have two sine squared X or sine of X is equal to one. This is a quadratic equation in sine and we have a sine X squared term. We have a sine X term and we have a constant term. So we're gonna move everything to one side two sine squared X plus sine X minus one is equal to zero. And we can do this in two ways. We recognize the factors from here. We can factor right away. If we're not recognizing the factors, we can go ahead and use the quadratic formula. So we'll do that and show that process in case you need it. So what we're gonna get is that sin X is going to be equal to negative B. So we can think about our coefficients for our quadratic equation. So we have A is going to be two, B is going to be one and C is going to be negative one and those coefficients of our quadratic equation. So we have Sinex is negative B which is gonna be negative one plus or minus the square root of B squared. So one squared minus four multiplied by A which is two multiplied by C which is negative one. We're gonna divide all of that by two A. So two multiplied by two. Let's go ahead and simplify inside of the square root. We have one squared which gives us one. We're subtracting four, multiplied by two multiplied by negative one. OK. So we have one minus negative eight, that's one plus eight. So we get nine. So we have a Sinex gonna be equal to negative one plus or minus the square root of nine divided by, for what we know that the square root of nine is equal to three. So sine of X is going to be equal to negative one less or minus three divided by four. And we get two solutions out of this, we get negative one and we get one. No, if SX is equal to negative one, we know that X is going to be equal to three pi divided by two, 270 degrees. And we know that sin is too pi periodic. So we can add any integer multiple of two pi to this. And we will still get that result. So X is gonna be three pi divided by two plus two pi N where N is an integer. So that's our first solution. Now we go to our second solution and our second solution sin X is equal to a half. We know that like this gives us X is equal to pi divided by six. And again, we can add two pi N to it because it is two pi periodic. We have to be a little bit careful with this solution. A sine X is equal to a half. We get pi divided by six as our reference angle. If we think about drawing this out on or methylene. OK. So we can think about the first one. Sinex is negative one. It's gonna lie over over here. Sorry, not there. Let me draw this over again. Sin X is equal to negative one lies along the negative x axis. So this angle that goes counter clockwise is gonna be three pi divided by two. So that's our first solution. Then we have a reference angle of pi divided by six in a lie up in quadrant, what we have this angle of pi divided by six. But we know that sign is also positive in quadrant two. OK. So if sin is also positive in quadrant two, and we can look at our reference angle as measured from the negative X axis. OK. So we're gonna measure this reference angle from the negative X axis which is pi divided by six. This is also going to be a solution. It also satisfies sine X is equal to one half. And so the angle that we get is gonna be measured from the positive X axis around. And we can see that that is gonna be pie minus pi divided by six because Pi makes that straight line and we're just going pi minus pi divided by six short of that. Sorry. So we get five pi divided by six. And so we had that first solution for our second case, we also have another solution. X can be five pi divided by six. And again, we can add two pi N because this is two pi period. And we can check these on our calculator. If you go ahead and substitute sign a five pi divided by six into your calculator, you're gonna get that value of a half. And so what we have is we actually have three sets of solutions comparing this to our answer choice. We can see that it matches with answer choice. So we have pi divided by six plus two N pi five pi divided by six plus two N pi and three pi divided by two plus two N pi or N in anger. Thanks everyone for watching. I hope this video help see you in the next one.

Key Concepts

Solving Quadratic Trigonometric Equations

General and Principal Solutions of Trigonometric Equations

Converting Between Radians and Degrees and Rounding

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