Determine the exact roots for each trigonometric equation or statement in the specified domain.

Determine the exact roots for each trigonometric equation or...

Key Concepts

Basic Trigonometric Equations

This concept involves solving equations where a trigonometric function is set equal to a specific value. The process typically includes isolating the trigonometric function, determining the general solution using inverse trigonometric functions or known values, and then refining the solution based on the given domain.

Domain and Periodicity

Trigonometric functions are periodic, meaning they repeat their values at regular intervals. When solving trigonometric equations, it is important to consider the specified interval (or domain) to determine which of the infinitely many solutions are valid within that range. The periodic nature of sine, cosine, and tangent functions allows us to find initial solutions and generate additional ones by incorporating full period rotations.

Reciprocal Functions and Undefined Values

Reciprocal trigonometric functions such as cosecant, secant, and cotangent are undefined where their corresponding base functions (sine, cosine, and tangent respectively) are zero. Recognizing these points of discontinuity is crucial when solving problems that involve determining where a particular function is undefined.

Trigonometric Equations Involving Quadratic Forms

Some trigonometric equations involve squared functions, forming a quadratic-type equation in a trigonometric function. Solving these equations typically involves isolating the squared term, applying algebraic techniques to remove the square (such as taking square roots, while considering both the positive and negative solutions), and then using trigonometric methods to solve for the angle.

Reference Angles and the Unit Circle

The unit circle is a fundamental tool in trigonometry that provides exact values for sine, cosine, tangent, and other functions at standard angles. Using reference angles helps in identifying the exact roots and corresponding angles that satisfy trigonometric equations by relating them to known values on the unit circle.

Transcript

00:03 All right, we want to find the exact roots for four equations that we've been given.

00:10 The first is equation 2, cosine of theta, minus radical 3 equals 0.

00:22 And we are looking at the interval where theta is between 0 and 2 pi.

00:36 So in other words, one trip around the unit circle on the coordinate axis.

00:41 Okay? so the first thing we're going to do is solve this equation so that we have the cosine of theta.

00:49 In other words, we're going to solve for cosine of theta.

00:52 So the first thing i'm going to do here is add radical 3 to both sides of this equation.

01:00 So 2, cosine of theta will equal radical 3.

01:06 Then if i divide both sides by 2, cosine of theta is now equal to radical 3 over 2.

01:15 Now there's a few things we have to pay attention to here.

01:18 The first is the fact that cosine is equal to a positive value.

01:24 And off to the side over here, we want to remember, when we're talking about the unit circle, right, we have our x and y coordinates.

01:33 I have 0, pi over 2, pi, 3 pi over 2, and 2 pi radians going all the way around.

01:47 And this would be the coordinate 1 .0.

01:49 0, 0, 1 at pi, x is going to be negative 1, y is going to be 0, and i'm going to have 0, negative 1.

02:09 So we want to remember, in quadrant 1, x, y are both positive, and quadrant 2, x is negative y is positive, and 3, they're both negative, and quadrant 4, x is positive, y is negative.

02:20 That's important because by definition, we know on the unit circle, cosine of theta is equal to the x coordinate.

02:31 Technically, cosine of theta is x over the radius, but on the unit circle, the radius is 1, so it's just equal to x.

02:38 So cosine is going to be positive where x is positive.

02:41 That means in this particular case, we're talking about quadrants 1 and 4.

02:52 In quadrant 1, this is easy.

02:53 Now, what we also have to remember is our values on the unit circle.

02:58 And we should know at this point that the cosine of pi over six radiance, which is equivalent to 30 degrees, is equal to radical 3 over 2.

03:15 Okay, that's just based on our unit circle.

03:17 Again, i'll throw that up here real quick.

03:20 There's certain values that we have to know that we should have memorized.

03:30 You have pi over 3, pi over 4, excuse me, that should have been pi over 6, and this is going to be pi over 3, which is equivalent to 30, 45, and 60 degrees.

03:48 And at each of these coordinates, x is going to be radical 3 over 2, y is 1⁄2, have 45, x and y are both radical 2 over 2.

04:01 Whoops, got a little too close, to that sidebar didn't i? x and y are both radical 2 over 2.

04:15 And at pi over 3 or 60 degrees, x is 1 half, x is 1 half, and y is radical 3 over 2.

04:44 Now i'm trying to squeeze that in.

04:46 The point is if cosine is equal to radical 3 over 2, theta is pi over 6 radiance.

04:52 Now that's in quadrant 1.

04:55 In quadrant 4, pi over 6 radiance serves as a reference angle.

05:04 When the that means then is theta in order to rotate to pi over six, theta is going to be two pi minus, right? we've gone almost all the way around except for pi over six.

05:21 So it's going to be two pi minus pi over six, which ends up becoming 11 pi over six radians.

05:33 So those are the two solutions to that particular equation.

05:37 Now, i talked about a lot of stuff here.

05:40 A lot of this is going to be the same ideas we're going to talk about going forward, so they should go a little bit quicker.

05:49 So the next one, it's kind of an odd one.

05:55 Kosikant of theta is undefined.

06:02 We want to know when co -sikin is undefined on the interval where theta is between 0 and 360 degrees.

06:11 So once again, we're talking about one trip around the unit circle.

06:15 Now, again, this is one of our identities.

06:17 We want to remember, cosecant is the reciprocal of the size.

06:25 Function.

06:26 So in other words, 1 over sine theta, this is going to be undefined wherever sign is equal to 0.

06:34 So once again, when we think about the unit circle that i have drawn above, i'm just going to do this real quick, at 0, we're at the coordinate 1 .0.

06:46 At 90, i'm going to be at the coordinate 0 .1.

06:52 At 180 degrees, this is the coordinate negative 1, 270.

06:57 270.

06:59 That's going to be 0 negative 1.

07:02 So when we want to talk about where sine is equal to 0, sine is equal to 0, wherever y is equal to 0, which is here at 0 degrees, and again here at 180 degrees.

07:15 So, cosecant will be undefined where sine is 0, or in other words, co -sicant is undefined when theta is equal to 0 or 180 degrees.

07:29 The next problem we want to look at is 5 minus the tangent square of theta equals 4.

07:50 And we are looking now on the interval between negative 180 degrees and between 180 and 360.

08:12 That's the interval we're looking at.

08:15 So once again, like the first one, i need to do whatever algebra is going to take to solve for tangent of theta.

08:23 So i will start by subtracting 5 from both sides, and i get negative tangent square theta equals negative 1.

08:35 And next, if i divide both sides by negative 1, then the tangent squared of theta equals 1.

08:46 And now at this point, if we take the square root of both sides, then the tangent of theta is equal to 1.

08:59 So once again, using what we know about the unit circle, tangents equal to 2...

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