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cosx-sinx=-1...

`cosx-sinx=-1`

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The correct Answer is:
`x=(2npi+(pi)/(2))orx=(2n-1)pi, "where "n in I`
`cosx-sinx=-1`
`rArr(1)/(sqrt(2))cosx-(1)/(sqrt(2))sinx=(-1)/(sqrt(2))`
`"cos"(pi)/(4)cosx-"sin"(pi)/(4)sinx=-"cos"(pi)/(4)=cos(pi-(pi)/(4))`
`rArrcos(x+(pi)/(4))="cos"(3pi)/(4)`
`rArrx+(pi)/(4)=2npi+-(3pi)/(4)`
`rArrx+(pi)/(4)=2npi+(3pi)/(4)orx+(pi)/(4)=2npi-(3pi)/(4)`
`rArrx=2npi+(pi)/(2)orx=(2n-1)pi" where"n in I`.
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