# pre-cal

posted by .

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!

## Similar Questions

1. ### trig

Verify the trigonometric identity: [(1–sin²x)/sin²x]–[(csc²x–1)/cos²x]= -tan²x I can't figure this out.
2. ### trig

Verify the trigonometric identity: [(1–sin²x)/sin²x]–[(csc²x–1)/cos²x]= -tan²x I still can't figure this out.
3. ### pre-calc

tanx= -12/5 x in Q 2. find sin2x= cos2x= tan2x=
4. ### pre-calculus

tanx= -12/5 x in quadrant 2 find; sin2x= cos2x= tan2x=

Which of the following are identities? Check all that apply. (Points : 2) sin2x = 1 - cos2x sin2x - cos2x = 1 tan2x = 1 + sec2x cot2x = csc2x - 1 Question 4. 4. Which of the following equations are identities?
6. ### Trig help

Can someone help me find ALL solutions in the interval [0,2π) for the given equations. Please show work. 1) sin2x = sin2x 2) tan2x +tanx = 0
7. ### Math

Solve the equation on the interval [0, 2π). tan2x sin x = tan2x would it be 0, pi/2 ?
8. ### maths

solve: tanx+tan2x+√3tanx tan2x=√3 (0<=x<=360)
9. ### math

sin2x − cos2x − tan2x = 2sin2x − 2sin4x − 1/ 1 − sin2x for x = 45
10. ### trig

secx = 4 in Q2 find sin2x,cos2x,tan2x

More Similar Questions