In Exercises 63–84, use an identity to solve each equation on the...

Question

In Exercises 63–84, use an identity to solve each equation on the interval [0, 2𝝅).sin x + cos x = 1

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Hello, today we're going to be solving the following trigonometric equation on the interval from 0 to 2 pi what we are given is sign of X minus cosine of X is equal to negative one. Now, since we are given the sine function and cosine function, let's go ahead and use the help of the trigonometric identities that involve sign and cosine a common trigonometric identity we can use is sine squared of X plus cosine squared of X is equal to one. In order to use this trigonometric identity, we'll first need to get the sine squared function and cosine squared function into our trigonometric equation. And we can do that by squaring both sides of the given equation. On the left hand side, sign of X minus cosine of X squared will allow us to rewrite the left hand side as sine squared of X minus two sign of X cosine of X plus cosine squared of X is equal to the right side, which is negative one squared, which will give us the value of positive one. Now we're going to reorder the terms on the left hand side if we were to put the sine squared and cosine squared quantity together we can rewrite the left hand side as sine squared of X plus cosine squared of X minus two sine of X cosine of X is equal to positive one. Now notice that we have sine squared of X plus cosine squared of X. On the left hand side of the equation, we can use the identity to rewrite this quantity as one and we can rewrite the left hand side as one minus two sine of X cosine of X equal to one. We can further simplify the equation by subtracting one to both sides, leaving us with negative two sine of X cosine of X equal to zero. And we can divide both sides of the equation by negative two to simplify the equation to sine of X cosine of X is equal to zero. Now, since the equation is equal to zero, we can set the factors of X equal to zero. And what we'll end up with is two separate equations. We have sin of X is equal to zero and cosine of X is equal to zero. Now, in order to solve for these equations, we need to ask ourselves the question, what angles of X can we plug in a sign to give a zero? And what angles of X can we plug in a cosine to give a zero? Well, in order to answer this question, we're going to need to use the unit circle. So let's go ahead and draw a unit circle. Now recall the problem only wants solutions of X on the range from 0 to 2 pi not including two pi. The four major angles of the unit circle is zero pi over two pi and three pi over two. Now the terminal point at zero is one comma zero. The terminal point at pi over two is zero, comma negative I'm sorry, positive one zero comma positive one. The terminal point at pi is negative one comma zero. And the terminal point at three pi over two is zero, comma negative one. Now recall that cosine is any X value of the terminal points and sine is any Y value of the terminal points. Let's go ahead and start with S sign of X is equal to zero. In order to answer this question, we need to look at the angles that have zero for the Y position of the terminal point. While there are two angles that have Y as zero, the angle of zero and the angle of pi have zero for the Y values. So that means the angles that we can plug into sine to give us the value of zero is going to be zero and pi so X is going to equal to zero and X is going to equal to pi. Now for cosine of X is equal to zero, we want to ask ourselves what terminal points have zero in the X position. While there are two angles that have zero in the X position. Those angles are pi over two and three pi over two as the exposition is going to be zero. So the angles that we can plug in a cosine to give us the value of zero is going to be pi over two and three pie over two. So what we currently have is four different solutions to X. But we'll need to verify that these solutions work with the original equation. So what we're going to need to do is we're going to need to plug in these values of X into the original equation to verify that there are solutions starting with X is equal to zero. If we were to plug this into the equation, we get S sign of zero minus cosine of zero is equal to negative one sine of zero will give us the value of zero and cosine of zero will give us the value of one on the left hand side, zero minus one will give us the value of negative one and we have negative one is equal to negative one, which is a true statement. So that means X is equal to zero is going to be our first solution of X. Next, let's check X is equal to pi. When X is equal to pi, we get sign of pi minus cosine of pi is equal to negative one sign of pi will give us the value of zero and cosine of pi will give us the value of negative one, negative, negative one will give us the value of positive one. And the left hand side will be zero plus one which gives us one. So we are left with one is equal to negative one, which is a false statement. That means X is equal to pi cannot be a solution for X. Next, let's check X is equal to pi over two. When X is equal to pi over two, we get sign of pi over two minus cosine of pi over two is equal to negative one sign of pi over two will give us the value of one and cosine pi over two will give us the value of zero. We have one minus zero on the left hand side which will leave us with one and we have one is equal to negative one which is a false statement. So X equal to pi over two cannot be a solution for X. Finally, let's check X is equal to three pi over two. Plugging this into the equation gives a sign of three pi over two minus cosine of three pi over two is equal to negative one. Sine of three pi over two will give us the value of negative one and cosine three pi over two will give us the value of zero that will leave us with negative one is equal to negative one, which is a true statement. So X equal to three pi over two will be a solution for X. And that means we have two total solutions for XX is equal to zero and X is equal to three pi over two. And with that being said, the answer to this problem is going to be a. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

In Exercises 63–84, use an identity to solve each equation on the interval [0, 2𝝅).sin x + cos x = 1

Key Concepts

Trigonometric Identities

Sine and Cosine Functions

Solving Trigonometric Equations

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