Match the following :
Column-I Equation | Column-II General Solutions |
(i) 2 sin θ – = 0 | [A] nπ + (–1)n |
(ii) 2 sin 2θ + | [B] 2nπ – = 2 sin θ + 2 cos θ |
(iii) sin 2θ + cos 2θ + 4 sin θ | [C] 2nπ + = 1 + 4 cos θ |
(iv) cos2θ = | [D] nπ + |
Text Solution
Verified by Experts(i) [A]; (ii) [A]; (iii) [D]; (iv) [A]
Ans.
(i) [A], [C]
(ii) [A], [B], [C]
(iii) [D]
(iv) [A], [B], [C]
Sol. (i) sin θ = 
= sin 
⇒ θ = n π + (–1) n 
= 2n π + 
(ii) 4 sin θ cos θ – 2 sin θ – 2 
cos θ + 
= 0
⇒ (2 sin θ – 
) (2 cos θ – 1) = 0
⇒ sin θ = 
⇒ θ = n π + (–1) n 
, cos θ = 
⇒ θ = 2n π ± 
(iii) sin 2 θ + cos 2 θ + 4 sin θ = 1 + 4 cos θ
⇒ 2 sin θ cos θ + 1 – 2 sin 2 θ + 4 sin θ = 1+ 4cos θ
⇒ 2 sin θ (cos θ – sin θ ) – 4(cos θ – sin θ ) = 0
⇒ (2 sin θ –4) (cos θ – sin θ ) = 0
⇒ sin θ = 2 or sin θ = cos θ
⇒ tan θ = 1 ⇒ θ = n π + π /4, n ∈ I
(iv) cos 2 θ = 1/4= cos 2 
⇒ θ = 2n π ± 
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