Trigonometric EquationHard
Question
Find all values of θ lying between 0 and 2π satisfying the equation r sin θ = √3 and r + 4 sin θ = 2(√3 + 1)
Options
A.
B.
C.
and 
and D.None of these
Solution
θ ∈ [0, 2π] equation rs in θ = √3 and r + 4 sin θ = 2(√3 + 1) we have to solve for θ
∴ We shall eliminate r from (i) and (ii) r =
from
+ 4sinθ = 2(√3 + 10
√3 + 4sin2θ = 2(√3 + 1) sin θ
⇒ 4sin2θ - 2√3 sin θ - 2 sinθ + √3 = 0
⇒ 2sinθ [2sinθ - √3] -1 [2sinθ - √3] = 0
⇒ (2sinθ - 1) (2sinθ - √3) = 0
⇒ sinθ =
or sinθ = 
⇒ θ =
or θ = 
∴ We shall eliminate r from (i) and (ii) r =
from
+ 4sinθ = 2(√3 + 10√3 + 4sin2θ = 2(√3 + 1) sin θ
⇒ 4sin2θ - 2√3 sin θ - 2 sinθ + √3 = 0
⇒ 2sinθ [2sinθ - √3] -1 [2sinθ - √3] = 0
⇒ (2sinθ - 1) (2sinθ - √3) = 0
⇒ sinθ =
or sinθ = ⇒ θ =
or θ = Create a free account to view solution
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