Solve `2 sin^(2) x-5 sin x cos x -8 cos^(2) x=-2`.

Explanation

If `cos x=0`, then
`2 sin^(2) x=-2 or sin^(2) x=-1`, which is not possible.
Clearly, `cos x ne 0`
Now, `2 sin^(2) x-5 sin x cos x -8 cos^(2) x=-2`
Dividing both sides by `cos^(2)x`, we get
`2 tan^(2) x-5 tan x-8=-2 sec^(2) x`
or `2 tan^(2) x-5 tan x-8+2 (1+ tan^(2) x)=0`
or `4 tan^(2) x-5 tan x-6=0`
or `(tan x-2) (4 tan x+3)=0`
Now, `tan x-2 =0`
Let `tan x=2=tan alpha`
`rArr x=npi +alpha = npi +tan^(-1) 2, n in Z`
or `4 tan x+3=0`
`rArr tan x=-3/4 = tan beta` (let)
`rArr x=mpi + tan^(-1) (-3/4)`, where `m in Z`