1.

Find the general solution of each of the equations : (i) `sin2x=-(1)/(2)` (ii) `tan3x=-1`

Answer» (i) `sin2x=-(1)/(2)=-"sin"(pi)/(6)=sin(pi+(pi)/(6))="sin "(7pi)/(6)`
`rArrsin2x="sin"(7pi)/(6)`
`rArr2x={npi+(-1)^(n)*(7pi)/(6)}` where `ninl`
`rArrx={(npi)/(2)+(-1)^(n)*(7pi)/(12)}`, where `ninl`.
Hence , the general solution is x `={(npi)/(2)+(-1)^(n)*(7pi)/(12)}` , where `ninl`.
(ii) `tan3x=-1=-"tan"(pi)/(4)=tan(pi-(pi)/(4))="tan"(3pi)/(4)`
`rArrtan3x" tan"(3pi)/(4)`
`rArr3x=(npi+(3pi)/(4))` where `ninl`.
`rArrx=((npi)/(3)+(pi)/(4))`, where `ninl`.
Hence , the general solution is `x=((npi)/(3)+(pi)/(4))` where `ninl`.


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