General solution of `sin^(2) x-5 sin x cos x -6 cos^(2)x=0` is

Options

A. `x=n pi-pi//4, n in Z` only
B. `npi+tan^(-1) 6, n in Z` only
C. both (a) and (2)
D. none of these

Answer

C

Explanation

Dividing the given equation by `cos^(2) x`, as `cos x =0` does not satisfy the equation, we have
`tan^(2) x-5 tan x -6 =0`
or `(tan x+1) (tan x-6)=0`
or `tan x=-1 or tan x=6`
If `tan x=-1=tan (-pi//4)`, then
`x=npi-pi//4, AA n in Z`
and, if `tan x=6=tan alpha` (say)
`rArr alpha = tan^(-1) 6`, then,
`x=n pi +alpha =n pi+tan^(-1) 6, AA n in Z`
Hence, `x=npi-(pi//4), n pi + tan^(-1) 6, n in Z`.