DeterminantHard
Question
The value of θ lying between 
and 0 ≤ A ≤ 
and satisfying the equation 
= 0 are -
and 0 ≤ A ≤ and satisfying the equation = 0 are -Options
A.A = 
, θ = - 
, θ = - B.A = 
= 0
= 0C.A = 
, θ = - 
, θ = - D.A = 
, θ = 
, θ = Solution
= 0 C1 → C1 + C2
⇒
= 0⇒
= 0⇒ 2(1 + 2sin4θ) - 2sin4θ = 0
⇒ 1 + sin4 θ = 0
sin4θ = - 1
4θ = -
or 4θ = 
θ = -
or θ = 
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