Column-I | Column-II |
(A)Number of solutions of sin2θ + 3 cos θ = 3 | (P)0 in [– π, π] |
(B)Number of solutions of sin x . tan 4x = cos x | (Q)1 in (0,π) |
(C)Number of solutions of equation | (R)4(1 – tan θ) (1+ tan θ)sec2θ + = 0 where θ∈ |
(D)If [sin x] + [cosx] =3, where x ∈ [0, 2π] | (s)5then [sin 2x] equals (Here [.] denotes G.I.F.) |
Text Solution
Verified by Experts1
→ (Q)
→ (S)
→ (R)
→ (P)
Sol. sin 2 θ + 3 cos θ = 3 ⇒ 1 – cos 2 θ + 3cos θ = 3 ⇒ cos 2 θ – 3cos θ + 2 = 0 ⇒ cos θ = 1, 2
⇒ cos θ = 1 ( cos θ ≠ 2) ⇒ θ = 0 in [– π , π ] No. of solution = 1
sin x . tan 4x = cos x ⇒ 
⇒ sin4x sinx – cos4x cosx = 0 ⇒ cos5x = 0
⇒ 5x = (2n + 1) π /2 ⇒ x = (2n + 1) π /10
⇒ x = 
, 
, 
, 
, 
in (0, π )
So there are five solutions.
(1 – tan 2 θ ) sec 2 θ + 
= 0
⇒ (1 – tan4 θ ) + 
= 0
⇒ (1 – x 2 ) + 2x = 0 where x = tan 2 θ
⇒ 2x = x 2 – 1 ⇒ x = 3
from graph number of solutions = 4
[sin x] + [ 
cosx] = – 3 ⇒ [sin x] = – 1 and [cosx] = – 2
⇒ π < x < 2 π and 
⇒ 
⇒ 
⇒ 
for 
; 
, 
∴ 
⇒ 
⇒ 0 < sin 2x < 1 ⇒ [sin2x] = 0
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