InterviewSolution
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InterviewSolution
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Solve : `secx-tan x=sqrt(3)`. |
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Answer» We have `secx-tanx=sqrt(3)rArr(1)/(cosx)-(sinx)/(cosx)=sqrt(3)rArr(1-sinx)=sqrt(3)cosx` `rArrsqrt(3)cosx+sinx=1`. Thus , the given equation becomes `sqrt(3)cosx+sinx=1`. Dividing both sides of (i) by `sqrt((sqrt(3))^(2)+1^(2))` , i .e by 2, we get `(sqrt(3))/(2)cosx+(1)/(2)sinx=(1)/(2)` `rArrcosx"cos"(pi)/(6)+sinx" sin "(pi)/(6)=(1)/(2)` `rArrcos(x-(pi)/(6))="cos"(pi)/(3)` `rArr(x-(pi)/(6))=2npi+-(pi)/(3), "where "n in I[becausecostheta=cosalpharArrtheta=2npi+-alpha]` `rArr(x-(pi)/(6))=(2npi+(pi)/(3))or(x-(pi)/(6))=(2npi-(pi)/(3))` `rArrx=2npi+(+(pi)/(3)+(pi)/(6))orx=2npi+(-(pi)/(3)+(pi)/(6))"where "ninI` `rArrx=(2npi+(pi)/(2))orx=(2npi-(pi)/(6)), "where "ninI` `rArrx=(2npi-(pi)/(6)), "where"n inI` `[because"secx is not defined when x"=(2npi+(pi)/(2))]`. Hence , the general solution is `x=(2npi-(pi)/(6)), "where"ninI`. |
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