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). sec² x - 2 = 0

Accepted Answer

Hey, everyone in this problem, we're given the following trigonometric equation. 10 minus five CO C squared X is equal to zero. And we're asked to use algebraic techniques to solve in the interval from 0 to 2 pi excluding two pi and to give the answer in radiance, we have four answer choices here. Option A is that there are no solution. And then option BC and D all have a set containing four different possible X values. So we're gonna start by writing out our equation, we have 10 minus five co C can't squared X is equal to zero. We're solving a trigger equation like this. We're gonna treat it just like any other equation. OK. So we want to get all of the numbers to one side and X on the other side, what we're gonna do is start by moving our five cos squared X to the right hand side by adding, we get 10 is equal to five. They can't square X. We're gonna divide both sides by five, we have 10, divided by five gives us two. And on the right hand side, we have co can squared X. I will recall that co can is equal to one divided by sine of X. So we can write Coy can square of X as one divided by sine squared X right. We are going to take the reciprocals on both sides. We get that sine squared of X is equal to one half. And if we take the square root, we get that sign of X is equal to plus or minus one divided by the square root of two. OK. And this is really, really important the plus or minus. When you take the square root, you have to remember that you get both roots and intrigue. This is gonna make a difference of having maybe just two solutions or four solutions. OK. So it's really important to remember both of those and then to evaluate which one makes sense in the context of the problem. OK. So we have sin of X which is equal to plus or minus one divided by the square root of two. So let's think about a special triangle that relates the side lengths, one and the square root of two. And recall the special triangle where the two side links are one. OK? Both side lengths are one. Then the hypotenuse is gonna be the square root of one squared plus one squared, which is the square root of two K. By we know that this is a 90 degree angle in our triangle. We've drawn a right angle triangle, both of the two smaller sides have the same length. And so both other angles in this triangle are gonna be pi divided by four, all right. So this is our special triangle. And these are really, really important. If you can remember your special triangles, you'll be able to remember or to know the angles of all of these different values. OK. So sin of X cosine of X tangent of X with all of these values without having to memorize each one of those separately. OK. So just learn and know your special triangles. Now we are gonna talk about the reference angle first. OK. So we're gonna call that X prime, that's gonna be an angle between zero and pi over two and that's measured from the X axis. So in this case, we're not worried about the positive or the negative, we're just worried about, we have one side as one. We have the hypotenuse is the square root of two. OK. We can see that the angle is gonna be pi over four from our special triangle. So our reference angle is gonna be pi divided by four. But we've said that this could be sign could be positive or negative. OK? If sine is positive or negative, that means it could be in any quadrant. OK. So sine positive or negative, that means we could have a solution in any quadrant. So let's go ahead and draw out all of our quadrants, draw out this reference angle in each one and see what solutions we get. So we have our Cartesian plane here. We're in a draw and we have a pi divided by four angle. And that's coming from the x axis. This is the positive X axis and quadrant run upwards in quarter two, we have that angle. But again, we always measure from an X axis. So in this case, we're going from the negative X axis up. Yeah, we get pie over four. In quadrant three, we do the same thing. We go from the negative X axis in this case, downwards pi divided by four. And finally, in quadrant four, we're going from the positive X axis downwards to get our angle of pi divided by four. So we have these reference angles and these four possible solutions shown on our diagram now and now we need to figure out what the actual angle is. So for the angle and quadrant one, OK. We measure again from the positive X axis. This is just gonna be pi over four for the other angles. OK. We always want to measure from the positive X axis in the counterclockwise direction. So we're gonna have to have different values to represent this. So for the angle in quadrant two, we're gonna go from the positive X axis in the counterclockwise direction until we hit an angle. OK. And this is gonna be what we call X two. Now X two, if you can imagine this blue line continuing all the way until the negative X axis. OK. So that we have an entire straight line, we know that the angle would be pi yes, we have pi but we're subtracting our reference angle of pi divided by four. So X two is gonna be pi minus pi divided by four, which gives us the three pi divided by four. Now we're gonna do the same thing this time with the angle and quadrant three, we start from the positive X axis again and we go all the way around counterclockwise to our angle. This one is X three similar to X two. If we went and stopped at the negative X axis, that would be a straight line, that would be 180 degrees or pi radiance. And but we're going to pi radiance and then we're going a little bit more and we're going this reference angle more. So for X three, we have pi plus pi divided by four, which gives us an angle of five pi divided by four. All right. And finally, we have this angle in the fourth quadrant. We're gonna do the same thing from the positive X axis counterclockwise all the way around till we hit that angle. Now, if we went all the way around our circle one time, that would be two P radiant, we're going almost all the way around, but we're stopping pi divided by four short. So what we have for X four here is we have two pi minus pi divided by four which gives us seven pi divided by four. OK. So we found that there were four solutions. OK. So we have X one is equal to pi divided by four, X two is equal to three pi divided by four, X three is equal to five pi divided by four and X four is equal to seven pi divided by four. If we go up to our answer choices and compare what we found, we can see that this corresponds with answer choice. 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). sec² x - 2 = 0

Key Concepts

Trigonometric Identities

Quadratic Form in Trigonometric Equations

Solving Trigonometric Equations on a Given Interval

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