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find all solutions in interval [0,2pi]

tan²x = -3/2 secx

This question may look tricky, but its actually quite simple.

First: rearrange the pythagorean identity tan²x+1=sin²x so that it reads tan²x=sin²x-1

Then replace the tan²x in the original problem with the sin²x-1 that you got from rearranging the pythagorean identity

Now the equation should read:

Next bring the (-3/2)secx to the other side of the equation to make it read:

sin²x+ (3/2)secx -1

Look familiar?

Yes! It is a quadratic. So you should set it equal to zero and factor.

I used the quadratic formula to get the two answers: -2 and .5

Then look on the unit circle and ask yourself: Where is the sec -2? Where is the sec .5?

Answer your questions by finding the points on the unit circle. Use the radians as your answers and don't forget to add 2(pi)n or (pi)n respectively

Let me know if you need more exact instructions


Im still stuck on the following steps i have no idea what im doing here..........sorry this is the section that ive had sooo much trouble in.

I would do it this way:

tan²x = -3/2 secx

sin²x/cos²x = -3/2cosx

multiply both sides by cosx

sin²x/cosx = -3/2 cross-multiply
2sin²x = -3cosx then by the Pythagorean identity
2(1-cos²x) + 3cosx = 0
2cos²x - 3cosx - 2 = 0
(2cosx+1)(cosx-2) = 0
so cosx = -1/2 or cosx = 2, the last is not possible because cosine is between -1 and 1

then cosx=-1/2 means the angle is in the second or third quadrants.
x = 120º or x = 300º


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