1.

Solve `cos3x+cosx-cos2x=0`

Answer» Given equations:
`cos3x+cosx-cos2x=0`,
or `2cos(3x+x)/(2)cos(3x-x)/(2)-cos2x=0,`
or `2cos2xcosx-cos2x=0`
or `cos2x.(2cosx-1)=0`,
`rArr cos2x=0` or `2cosx-1=0`
If `cos2x=0`, then `2x=(2n+1)pi/2 rArr x=(2n+1)pi/4` where `n in z`
If `2cosx-1=0`, then `cosx=1/2 rArr cosx=cospi/3`
`rArr x=2npi+-pi/3`
Therefore, the general solution of given equation is `x=(2n+1)pi/2`.
or `2npi+- pi/3, n in Z` Ans.


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