Question

I need help on this question as I've been stuck on it for quite a while.! also the range is in radians from 0 TO PI. solve for x? thank you!!

`tan^2(2x-pi/6)=11-sec(2x-pi/6)`

Answer 1

See below.

Explanation:

For simplicity.

Let `x=(2x-pi/6)`

`tan^2x=11-secx`

`tan^2x+secx=11`

Using identities:

`color(red)bb(tanx=sinx/cosx)`

`color(red)(bb(secx=1/cosx)`

`sin^2x/cos^2x+1/cosx=11`

Add fractions:

`(sin^2x+cosx)/cos^2x=11`

`sin^2x+cosx=11cos^2x`

Identity:

`color(red)bb(sin^2x+cos^2=1)`

`1-cos^2x+cosx=11cos^2x`

Rearrange:

`-12cos^2x+cosx+1=0`

Multiply by `-1`

`12cos^2x-cosx-1=0`

Let `u=cosx`

`12u^2-u-1`

Factor:

`(4u+1)(3u-1)=0=>u=-1/4 and u=1/3`

`u=cosx=>cosx= -1/4 and 1/3`

`x=(2x-pi/6)=cos(2x-pi/6)=-1/4 and 1/3`

`2x-pi/6=arccos(cos(2x-pi/6))=arccos(-1/4)`

`2x-pi/6=arccos(-1/4)`

`x=(arccos(-1/4)+pi/6)/2~~1.17354` radians 5 .d.p.

`x=(-arccos(-1/4)+pi/6)/2+pi~~2.49165` radians 5 .d.p.

`2x-pi/6=arccos(cos(2x-pi/6))=arccos(1/3)`

`2x-pi/6=arccos(1/3)`

`x=(arccos(1/3)+pi/6)/2~~0.87728` radians 5 .d.p.

`x=(-arccos(1/3)+pi/6)/2+pi~~2.78791` radians 5 .d.p.

Answer 2

`x = pi/12 - 1/2 cos^-1 (1/3) + pi approx 2.788`
`x = pi/12 + 1/2 cos^-1 (1/3)approx 0.877`
`x = pi/12 - 1/2 cos^-1 (-1/4) + pi approx 2.492`
`x = pi/12 + 1/2 cos^-1 (-1/4) approx 1.174`

Explanation:

This seems like a problem that needs to be solved with a graphing calculator - the actual process is quite involved. I describe it here for reference.

`tan^2(2x-pi/6)=11-sec(2x-pi/6)`

We begin by moving everything to one side.

`tan^2(2x-pi/6)+ sec(2x-pi/6) - 11 = 0`

We substitute using the identity `tan^2(x) = sec^2(x) - 1`

`sec^2(2x-pi/6) + sec(2x-pi/6) - 12 = 0`

This can be factored into:

`(sec(2x-pi/6)-3)(sec(2x-pi/6)+4) = 0`

Which we split into two equations.

`sec(2x-pi/6)-3 = 0`
`sec(2x-pi/6)+4 = 0`

I will now continue with the simplification, which is relatively straightforward.

`sec(2x-pi/6)-3 = 0`
`sec(2x-pi/6) = 3`
`cos(2x-pi/6) = 1/3`
`2x - pi/6 = 2pi n_1 + cos^-1 (1/3)` or `2x - pi/6 = 2pi n_2 - cos^-1 (1/3)`
`x = pi/12 - 1/2 cos^-1 (1/3) + pi n_1` or `x = pi/12 + 1/2 cos^-1 (1/3) + pi n_2`

and

`sec(2x-pi/6)+4 = 0`
`sec(2x-pi/6) = -4`
`cos(2x-pi/6) = -1/4`
`2x - pi/6 = 2pi n_3 + cos^-1 (-1/4)` or `2x-pi/6 = 2pi n_4 - cos^-1 (-1/4)`
`x = pi/12 - 1/2 cos^-1 (-1/4) + pi n_3` or `x = pi/12 + 1/2 cos^-1 (-1/4) + pi n_4`

Since `0 <= x <= pi`, we look for values of `n_1`, `n_2`, `n_3`, and `n_4` (note: must be integers) to meet this constraint. I personally plugged in -1, 0, and 1 to see which would work.

We find that:
`n_1 = 1`
`n_2 = 0`
`n_3 = 1`
`n_4 = 0`
for `x` to meet its constraint.

Finally, we get our answer:

`x = pi/12 - 1/2 cos^-1 (1/3) + pi approx 2.788`
`x = pi/12 + 1/2 cos^-1 (1/3)approx 0.877`
`x = pi/12 - 1/2 cos^-1 (-1/4) + pi approx 2.492`
`x = pi/12 + 1/2 cos^-1 (-1/4) approx 1.174`
`square`

EDIT: we can verify our result graphically, using Wolfram Mathematica. We see that there are clearly 4 intersections.