Solve each equation algebraically over the domain 0^∘ ≤x<360^∘ . Verify your solution graphicall

Solve each equation algebraically over the domain 0^∘...

Key Concepts

Graphical Verification

Graphical methods provide a visual confirmation of algebraically derived solutions by plotting the functions or equations involved. This approach serves as a check to ensure that the computed solutions are indeed correct and helps in understanding the behavior of trigonometric functions over their domains.

Periodic Nature and Domain Considerations

Trigonometric functions are periodic, meaning their values repeat over regular intervals. Recognizing and using the period of the function is vital for determining all solutions within a specific domain. This knowledge ensures that one accounts for every possible angle that satisfies the equation in the given interval.

Trigonometric Identities

Trigonometric identities, such as the double angle formulas, are essential tools for rewriting and simplifying trigonometric equations. By expressing functions like cos 2x or sin²x in terms of cos x or sin x, one can reduce equations to forms that are easier to manipulate algebraically, making it possible to isolate the variable and solve for it.

Algebraic Manipulation and Factoring

Many trigonometric equations can be rearranged into algebraic forms or even quadratic equations in terms of a single trigonometric function. Techniques such as factoring common terms, applying the zero-product property, or using the quadratic formula are crucial for isolating the trigonometric function and finding its solutions.

Transcript

00:02 Okay, so we have all these functions here, four equations, and we're going to solve each algebraically.

00:08 So let's start with this first one.

00:10 We have this cosine 2x here, so we're going to need the double angle formula.

00:15 I've written that up here, right here, and we have this in terms of cosine, so let's see if we can get our double angle formula, which is this, also in terms of just cosine.

00:31 Okay, how can we do that? we want to use our fundamental identity, sine squared plus cosine squared equals 1.

00:40 And here, we're going to solve for our sine squared, and then plug this in here.

00:51 So, cosine 2x will be equal to cosine squared x minus 1 minus cosine squared x in parentheses.

01:02 So equal to, if we distribute this negative, we'll end up with a positive cosine squared.

01:09 So we add those and we get two cosine squared x minus one.

01:14 So this turns into cosine x minus two cosine squared x minus two cosine squared x minus 1 equals zero.

01:24 So to make this simple, i'm just going to move everything over to this side so that this two cosine squared x is positive.

01:31 So we'll end up with two cosine squared x minus cosine x minus 1 equals 0.

01:40 Okay, then let's solve it from there.

01:45 We're going to need to factor this.

01:48 How do we factor it? first, i'm going to, one second, i'll move this piece a little further down, so we more space.

01:57 Okay, how are we going to factor this? well, just think of the cosine square.

02:04 As an x.

02:06 So 2x squared minus x minus 1 equals 0.

02:13 If we factor this, we'll end up with this.

02:18 So you can forward that out to double check it.

02:22 If we just put in a cosine x instead of an x, we'll end up with plus 1 here, sine x minus 1.

02:30 Okay, and since that's equal to 0, we can just take each individual piece and set it equal to zero like this.

02:40 And the reason we can do that is because these are independent factors.

02:46 So if one of these is equal to zero, the whole thing is set equal to zero.

02:51 So we'll solve this one and we'll get to cosine x equal to negative one, cosine x equals negative one -half.

03:04 And we think of where cosine x could be equal to next.

03:07 1 1 half because of our all students take calculus is going to be the second or the third quadrant and because cosine refers to the x value and it's short we know that it's going to be two pi thirds and four pi thirds okay then cosine x equals one we just add that one over and we get x equals x x we get x equals 0.

03:45 Okay, and we have the domain 0 is less than or equal to x is less than 360.

03:52 So let's just, we can change this to degrees if we want to.

03:57 2 pi thirds is 120 degrees, 4 pi thirds is 2 times that, so 240 degrees.

04:06 Zero is just 0 degrees.

04:09 So there we go.

04:11 Let's solve the next one.

04:18 Okay.

04:20 Squared x minus 3, sine x equals 4.

04:24 So we're going to treat this like a quadratic again.

04:27 Our first step is always to set it equal to 0.

04:33 And if we factor this like x squared minus 3x minus 4, we end up with x minus 4 x plus 1.

04:44 So let's just stick that in for sine.