Solve each equation (x in radians and θ in degrees) for a...

Question

Solve each equation (x in radians and θ in degrees) for all exact solutions where appropriate. Round approximate answers in radians to four decimal places and approximate answers in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures.

3 csc² x/2 = 2 sec x

3 csc² x/2 = 2 sec x

Accepted Answer

Find all solutions of the given equation express your answer in radiance using the least possible nonne angle measure. We round your answer to the nearest 100th if necessary. Where our equation is five second squared X divided by two equals four second X and we have four possible answers. All being pairs of angle measures in terms of radiance. Now to solve this, let's first try and find what exits. We have five seconds squared X divided by two equals four second of X. Now because we know that sin is a reciprocal of cosine, we can rewrite this as five multiplied by one divided by cosine squared X divided by two equals four seeking the V. If we multiply both sides by cosine squared X divided by two, we end up giving the equation five equals four. See X cosine squared X divided by two. Now we'll have to make use of another trig identity. We should know one identity. We know is that coun of X divided by two is equals to plus or minus one plus cosine X divided by two, which is our half angle identity. And that is all underneath a square root. We can then rewrite our cosine squared as Kan squared X divided by two equals one plus cosine X divided by two. Now let's replace this back in our main equation, we have five equals four, see one plus cosine X divided by two. Now, from here, we just need to simplify five equals two second X one plus cosine X divided by two. And that too is now gone since we divided it into the four. So we will actually erase that. And now we can go ahead and distribute, we get five equals two, multiplied by seek an X plus, seek an X cosine X that gives us five equals two, multiplied by one divided by cosine X plus one. Since sin X and cosine X are reciprocal that will turn into one. Now, if we distribute our two, we get five equals two divided by cosine X plus two. And now we just need to multiply both sides by cosine to get rid of it. In the fraction. We end up giving five cosine X equals two plus two, cosine X. We can now subtract two cosine X from both sides. If I go over here, we end up getting five cosine X minus two cosine X which turns into three cosine X is equals to two. And finally dividing both sides by three gives us cosine X equals two dits. Now, this value is not on the unit circle. So we'll have to use a calculator to determine what X is. If we use X equals cosine inverse two thirds, we end up giving two values X equals 0.84 radiance and X equals to pi minus 0.84 radiance which is 5.44. Now, because these are angle measures on a potential unit circle and we loop around the circle for every two pi revolution, we will add two pi in to both equations. We can't simplify any farther. So we have two answers, 5.44 plus two pi N and 0.84 plus two pi N. This means the answer to our problem. It's answer A OK. I hope to help you solve the problem. Thank you for watching. Goodbye.

Key Concepts

Trigonometric Identities

Solving Trigonometric Equations

Angle Measures and Conversion

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