Question

For `n in Z ,` the general solution of `(sqrt(3)-1)sintheta+(sqrt(3)+1)costheta=2i s(n in Z)` `theta=2npi+-pi/4+pi/(12)` `theta=npi+(-1)^npi/4+pi/(12)` `theta=2npi+-pi/4` `theta=npi+(-1)^npi/4-pi/(12)`
A. `theta=2npi pm pi/4+pi/12`
B. `theta=n pi +(-1)^(n) pi/4+pi/12`
C. `theta=2npi pm pi/4`
D. `theta=npi +(-1)^(n) pi/4- pi/12`

Answer 1

Correct Answer - A
`(sqrt(3)-1) sin theta+(sqrt(3)+1) cos theta=2`
or `((sqrt(3)-1))/(2sqrt(2)) sin theta+ ((sqrt(3)+1)/(2sqrt(2))) cos theta=1/sqrt(2)`
or `"sin" pi/12 sin theta+"cos"pi/12 cos theta= "cos"pi/4`
or `cos(theta-pi/12)="cos" pi/4`
`rArr theta-pi/12=2npi pm pi/4, n in Z`
or `theta=2npi pm pi/4+ pi/12, n in Z`