(i) Given : sin `x=(1)/(sqrt(2))`. The least value of x in `[0, 2pi[" for which sin"x=(1)/(sqrt(2))is x = (pi)/(4)`.
`therefore"sin x = sin"(pi)/(4)rArrx=npi+(-1)^(n)*(pi)/(4)`, where ` n in I`.
Hence , the general solution is x = `npi+(-1)^(n)*(pi)/(4)`, where ` n in I`.
(ii) Given `cos x =(1)/(2)`.
The least value of x in `[0, 2pi["for which cos "x=(1)/(2) is x=(pi)/(3)`.
`therefore" cos x cos "(pi)/(3)rArrx=(2nx+-(pi)/(3))`, where ` n in I`.
Hence , the general solution is `x=(2nx+-(pi)/(3))`, where ` n in I`
(iii) Given : `tanx=(1)/(sqrt(3))`.
The least value of x in `[0 , 2pi[" for which tan "x=(1)/(sqrt(3))is (pi)/(6)`.
`therefore tanx ="tan "(pi)/(6)rArrx=(npi+(pi)/(6))`, where `n in I` .
Hence , the general solution is `x=(npi+(pi)/(6))`, where `ninI`.