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Questions  

The most general values of θ  satisfying the equation (1+2sinθ)2+(3tanθ1)2=0   are given by

a
nπ+π6,n∈z
b
nπ2+(−1)n7π6
c
2nπ+7π6,n∈z
d
2nπ+11π4,n∈z

detailed solution

Correct option is C

sum of squares =0 ⇒each term=01+2 sinθ=0  ⇒sinθ=−12,          3tanθ−1=0 ⇒ tanθ=13 ∴ θ∈θ3 then θ = 2nπ+7π6

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