Solve each equation in x over the interval [0, 2π) and each eq...

Question

Solve each equation in x over the interval [0, 2π) and each equation in θ over the interval [0°, 360°). Give exact solutions.

8 sec² x/2 = 4

Accepted Answer

Find the exact solutions of the given equation over the interval from 0 to 2. Pi where our equation is three seconds squared X divided by three equals six. And we have four possible answers. We have four different sets with different angles involving pi divided by four. Now to solve this, let's first rewrite our equation three seconds squared X divided by three equals six. And we can first divide both sides by three, we get second squared X divided by three equals two. Now something to note is that see it is the reciprocal function in particular. See here is one divided by cosine. So we can rewrite this s cosine squared X divided by three. Now the two on the right side, since it's the reciprocal turns into one half. Now let's also take the square root of both sides. This will get rid of the squared on the Kent where we end up with cosine X divided by three equals the square root of two divided by two. When simplified. And this has a positive and a negative value. Now let's look on our unit circle. But where we have cosine is equivalent to square root of two divided by two. If we look at a unit circle, the first place this happens is that pi divided by four, you write this as the W pi divided by four and it also occurs that's seven pi divided by four. Now for the negative, these occur at three P divided by four and five pab by four. So we have four angles we need to check. But now set our result equals to X divided by three. We can X divided by three equals pi divided by four, X divided by three equals three pi divided by four, X divided by three equals five pi divided by four NX divided by three equals seven pi divided by four. Now we also need to account for the looping behavior sign or cosine in our case because cosine and sine both loop. And by that, we mean that for every two pi revolution around the unit circle, the angle ends up the same. In other words, we can write this as pi divided by four plus two pi N. And we can put that for everyone of our requests. Now we can multiply three for all these equations. Since we have X divided by three, we end up getting X equals three pi divided by four plus six pie in X equals nine pi divided by four plus six pi in 15 pi divided by four plus six PN and finally 21 pi divided by four equals six PN. Now three of our values here or answers, we can knock off right off the bat because we notice those are larger than our interval. You know that our last three equations all end up being greater than two P. So these are all outset of our interval, which means we just look at X equals three pi divided by four plus six pi in if we let N equals zero, Rx will be three pi divided by four plus six pi multiplied by zero, which is just three pi divided by four. Now, if we try another index, we would notice that we're adding six pi which is larger than two pi, which means we have no other angles, which means the answer to our problem. Three pi divided by four. It's in T OK. Hope that help you solve the problem. Thank you for watching. Get that.

Solve each equation in x over the interval [0, 2π) and each equation in θ over the interval [0°, 360°). Give exact solutions.
8 sec² x/2 = 4

Key Concepts

Trigonometric Functions and Their Definitions

Solving Trigonometric Equations

Interval and Domain Considerations for Solutions

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