# pre-cal

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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

sin²x-1=(-3/2)secx

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?

Let me know if you need more exact instructions

Cori

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

Done!