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).                        __                      √ 2sin x cos x = --------                         4

Accepted Answer

Hello, today we're going to be solving the trigonometric equation on the interval from 0 to pi. So what we are given is sign of X multiplied by cosine of X is equal to the square root of 3/4. Now, one thing to note is that we are given a product of sign and cosine on the left hand side of the equation, there is a trigonometric identity that involves the product of sine and cosine. And that trigonometric identity is given to us as sine two, X is equal to two multiplied by sine X multiplied by cosine of X. In this identity, we have a product of sign and cosine. But this product has a coefficient of two. In order to use this identity, we'll need to get that coefficient of two on the left hand side of the equation. And in order to do that, we can multiply the entire the equation by two, multiplying the equation by two allows us to rewrite the equation as two multiplied by sin of X multiplied by cosine of X is equal to the square root 3/4 multiplied by two, which will give us the square root of 3/2. Notice that we have two sine of X cosine of X. On the left hand side of the equation, we can use our trigonometric identity to rewrite the left-hand side as sine of two X. So the new equation that we have is sine of two X is equal to the square root of 3/2. In order to solve this equation, let's use the substitution. We're going to allow Y to equal to two X. And if we apply this substitution, we can rewrite the equation as sign of Y is equal to the square root of 3/2. Now, in order to answer this question, we need to ask ourselves, what angles of why can we plug in a sign to give us the value of the square root 3/2? Well, in order to answer this question, we're going to need to use a unit circle. Now recall that any terminal point on the unit circle is defined as cosine comma sign cosine is every X value and sin is every Y value. So what angles have the value of square root 3/2 in the Y position of the terminal point? Well, there are two angles that contain this value. The first one is in the first quadrant of the unit circle at the angle pi over three, the terminal point at this position is one half comma square root 3/2. What this means? Is that if we were to plug in the angle pi over three into sine, we'll get a return value of the square root 3/2. So that means Y is going to equal to pi over three. Now the second angle that has the square with 3/2 in the Y position of the terminal point exists in the second quadrant. That angle is two pi over three and the terminal point at this angle is negative 1/2 comma square root 3/2. What this means is that if we plug in the angle two pi over three into sine, we'll get the value of the square over 3/2. So why is equal to two pi over three? Now keep in mind that we originally wanted to solve for X. So if we rebs tote Y is equal to two X back in these in these two solutions, we get two X is equal to pi over three and we get two X is equal to two pi over three. And keep in mind that we want all of the solutions in the range of 0 to pi not including two pi. Normally, what we'll need to do in order to get the rest of the solutions is we'll need to add two pie to the angles that we found. However, since we are solving for X, we need to first isolate X. And in order to do that, we can divide both sides of the equation by two that will leave us with X is equal to pi over six plus pi. And that will also give us a second value X is equal to two pi over six plus pi and two pi over six can be reduced to pi over three. What this means is that the first solutions of X is going to be pi over six and pi over three. But in order to get the remaining solutions of X, we'll need to add pi to both of these quantities. So let's go ahead and start with the first equation pi over six plus pi while we can go ahead and add these if we were to rewrite pi as six pi over six. So we have pi over six plus six pi over six and adding these quantities together will give us a final angle of seven pi over six. This is going to be the next solution of X from the range of 0 to 2 pi. And on the right hand side, we're going to add pie over three with pie. We can rewrite pi as three pi over three. And we get pi over three plus three pi over three and adding these quantities together will give us a final value of four pi over three. This is going to be the final solution of X within 0 to 2 pi. So just to summarize, we have four solutions in total, we have pi over six pi over 37 pi over six and four pi over three. 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). __ √ 2sin x cos x = -------- 4

Key Concepts

Trigonometric Identities

Double Angle Formulas

Interval Notation

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