Answer

`a rarr r, b rarr s, c rarr p, d rarr q`.

Explanation

a. `cos^(2) 2x- sin^(2) x=0`
`rArr cos 3x cos x=0`
`rArr cos 3x=0 or cos x =0`
`rArr 3x=(2n-1) pi/2 or x=(2n-1) pi/2, n in Z`
or `x=(2n-1) pi/6 or x=(2n-1) pi/2`
Hence, the general solution is `(2n-1) pi/6` as `(2n-1) pi/2` is
contained in `(2n-1) pi/6`.
b. `cos x+sqrt(3) sin x=sqrt(3)`
or `(cos x)/2+sqrt(3)/2 sin x=sqrt(3)/2`
or `cos(x-pi/3)=cos pi/6`
`rArr x-pi/3=2n pi pm pi/6, n in Z`
`rArr x=2npi +pi/2 or x=2n pi +pi/6`
c. `sqrt(3) tan^(2) x-(sqrt(3)+1) tan x+1=0`
or `sqrt(3) tan x(tan x-1)-(tan x-1)=0`
or `(tan x-1) (sqrt(3) tan x-1)=0`
`:. tan x=1 rArr x=npi +pi/4, n in Z`
or `tan x=1/sqrt(3) rArr x=n pi +pi/6, n in Z`
d. `tan 3x-tan 2x-tan x=0`
or `tan x tan 2x tan 3x =0`
`x=npi or n pi//2` (rejected) or `n pi//3`
Therefore, the general solution is `n pi//3` as `n pi` is contained in `n pi//3`.