Solve `4 cos theta-3` sec `theta=tan theta`.

Explanation

We have `4 cos theta-3 sec theta = tan theta`
or `4 cos theta-3/(cos theta)= (sin theta)/(cos theta)`
or `4 cos^(2) theta-3= sin theta`
or `4(1- sin^(2) theta)-3= sin theta`
or `4 sin^(2) theta + sin theta-1=0`
or `sin theta=(-1 pm sqrt(1+16))/8=(-1 pm sqrt(17))/8`
Now, `sin theta=(-+sqrt(17))/8`. Thus,
`sin theta=sin alpha`, where `sin alpha=(-1+sqrt(17))/8`
`rArr theta=n pi +(-1)^(n) alpha`,
where `sin alpha =(-1 + sqrt(17))/8 and n in Z`
Also, `sin theta=(-1-sqrt(17))/8`
`rArr sin theta= sin beta`, where `sin beta=(-1-sqrt(17))/8`
`rArr theta=n pi +(-1)^(n) beta, n in Z` where `sin beta =(-1-sqrt(17))/8`