(i) `cos4x=cos2x`
` rArr cos 4x-cos 2x=0`
`rArr-2" sin "((4x+2x))/(2)"sin"((4x-2x))/(2)=0`
`[becausecosC-cosD=-2"sin"((C+D))/(2)"sin"((C-D))/(2)]`
`rArr-2sin3xsinx=0`
`rArrsin
3x=0orsinx =0`
`rArr3x=npiorx=mpi`, where m, `ninI`
`rArrx=(npi)/(3)orx=mpi` , where m ,n `in` I.
Hence , the general solution is x `=(npi)/(3)orx=mpi` , where m ,n `in`I.
(ii) `cos3x=cos((pi)/(2)-2x)`
`rArr3x=2npi+-((pi)/(2)-2x)[costheta =cos alpha rArr theta = 2 npi +- alpha]`
`rArr3x=2npi+((pi)/(2)-2x)or3x=2npi-((pi)/(2)-2x)`, where `ninI`
`rArr5x=2npi+(pi)/(2)orx=(2npi-(pi)/(2))`,where `ninI`
`rArrx=((2npi)/(5)+(pi)/(10))orx=(2npi-(pi)/(2))` , where `ninI`.
Hence , the general solution is x `=((2npi)/(5)+(pi)/(10))orx=(2npi-(pi)/(2))`, where `ninI`.
(iii) `sin3x+cos2x=0`
`rArrcos2x=-sin3x=cos((pi)/(2)+3x)`
`rArrcos2x=cos((pi)/(2)+3x)`
`rArr2x=2npi+-((pi)/(2)+3x)`, where `ninI`
`rArr2x=2npi+((pi)/(2)+3x)or2x=2npi-((pi)/(2)+3x)` , where `ninI`
`rArrx=(-2npi-(pi)/(2))orx=((2npi)/(5)-(pi)/(10))` , where `n inI`
`rArrx=(2npi-(pi)/(2))orx=((2npi)/(5)-(pi)/(10))` , where `ninI`.
Hence , the general solution is x `=(2npi-(pi)/(2)orx=((2npi)/(5)-(pi)/(10))`, where `ninI`.
Note `(-2npi-(pi)/(2))and(2npi-(pi)/(2))` give the same result as `ninI`.
`sinmx+sinnx=0`
`rArr2"sin"((m+n)x)/(2)"cos"((m-n)x)/(2)=0`
or `"sin"((m+n)x)/(2)=ppior((m-n)x)/(2)=(2q+1)(pi)/(2)`, where p , q `in` I
`rArrx=(2ppi)/((m+n))orx=((2q+1)pi)/((m-n))`, where p , q `in` I
Hence , the general solution is `x=(2ppi)/((m+n))orx=((2q+1)pi)/((m-n))`, where p , q `in` I.
