precal
posted by ethan .
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=sin²x1
Then replace the tan²x in the original problem with the sin²x1 that you got from rearranging the pythagorean identity
Now the equation should read:
sin²x1=(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?
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
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 crossmultiply
2sin²x = 3cosx then by the Pythagorean identity
2(1cos²x) + 3cosx = 0
2cos²x  3cosx  2 = 0
(2cosx+1)(cosx2) = 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!
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