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Question

Solve the following equations :(i) sin θ+cos θ=2(ii) 3 cos θ+sin θ=1(iii) sin θ+cos θ=1(iv) cosec θ=1+cos θ(v) (31)cos θ+(3+1)sin θ=2

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Solution

(i) sin θ+cos θ=2We have,sin θ+cos θ=212 sin θ+12cosθ=1sinπ4sin θ+cosπ4cos θ=1 [cosπ4=sinπ4=12]cos(θπ4)=cos 0θπ4=2nπ,nzθ=2nπ+π4,nzθ=(8n+1)π4,nz(ii) 3 cos θ+sin θ=1Dividing both side by 2,we get32cos θ+12sin θ=12cos π6 cos θ+sin π6sin θ=12 [sin=π6=12,cosπ6=32]cos(θπ6)=cos π3θ=π6=2n±π3,nzθ=2nπ ±π3+π6,nzθ=(4n+1)π2Or(12m1)π6,nmz(iii) sin θ+cos θ=1We have,sin θ+cos θ=1Divide both side by2,we get12 sin θ+ 12 cos θ=12sin π4 sin θ+cos π4 cos θ=12 θ( θπ4)=cos π4 θ=π4=2nπ± π4,nZ θ=2nπ+ π2,2nπ nZ(iv) cosec θ=1+cos θWe have,cosec θ=1+cos θ 1sinθ=1+cosθsinθ 1=sin θ+cos θDivide both side by2,we get12 sin θ+12 cos θ=12sin π4 sin θ+cos π4 cos θ=12cos ( θ π4)=cos π4 θ=π4=2nπ± π4,nZ θ(2nπ+π2) Or 2nπ,nZ(v) (31)cos θ+(3+1)sin θ=2Divide on both sides by22(31)22 cos θ+(3+1)22 sin θ=12sin (θ+tan1(313+1))=sin π4θ=2nπ+π3 Or 2nπ π6 nZ


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