JEE Advanced (Single Correct MCQs): Trigonometric Functions & Equations - NDA MCQ

II quad or IV quad

But   

The given equation is

where  

LHS  = 

Q 1 + cos x < 2 and sin2 x £ 1 for 

And  R.H.S. =   

given equation is not possible for any real  value of x.

sin x + cos x = 1

(the set of integers)

where n = 0, ± 1, ± 2, . . . .

The given expression is

The given equation is sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x
⇒ 2 sin 2x cos x – 3 sin 2x = 2 cos 2x cos x – 3 cos 2x
⇒ sin 2x (2 cos x – 3) = cos 2x (2 cos x – 3)
⇒ sin 2x = cos 2x (as cos x ≠ 3/2)

⇒ tan 2x = 1 ⇒ 2x = nπ + π/4 ⇒

The given equation is (cos p – 1) x² + (cos p) x + sin p = 0 For this equation to have real roots D≥ 0
⇒ cos² p – 4 sin p (cos p – 1)≥ 0
⇒ cos² p – 4 sin p cos p + 4 sin² p + 4 sin p – 4 sin² p≥ 0

⇒ (cos p – 2 sin p)² + 4 sin p (1– sin p)≥ 0
For ever y real value of p (cosp – 2sinp)²≥ 0 and 1– sin p ≥ 0

The given eq is, tan x + sec x = 2 cos x

⇒ sin x + 1 = 2 cos² x ⇒ 2 sin² x + sin x – 1 = 0

⇒ (2 sin x –1)  (sin x +1 ) = 0 ⇒

But for given eq. is not defined,

∴ only 2 solutions.

For n = 2 the given equation is not satisfied.
Considering n > 1 and n ≠ 2

3 (sin x – cos x)⁴ + 6 (sin x + cos x)² + 4 (sin⁶ x + cos⁶ x)
=  3 (1 – sin ²x)² + 6 (1 + sin ²x) + 4[(sin² x + cos2x)³ – 3 sin² x cos²x (sin² x + cos² x)]

=  3 – 6 sin 2x + 3 sin² 2x + 6 + 6 sin 2x

= 13 + 3 sin² 2x – 3 sin² 2x = 13

The given equation is  2 sin² θ- 3 sinθ- 2 = 0

⇒ (2 sinθ + 1) (sinθ – 2) = 0

⇒ sinθ      [∴ sinθ - 2=0 is not possible]

⇒ sinθ = sin(-π/ 6) = sin (7π/ 6)

⇒ θ= nπ + (-1)ⁿ(- π/6) =np + (-1)ⁿ7π/6

⇒ Thus, θ = nπ + (-1)ⁿ7π/ 6

We have

But 

⇒ x – y = 0 [ as perfect square of real number can never be negative] Also then x ≠ 0 as then sec² θ will become indeterminate.

Given that in 

Also tan P/2 and tan Q/2 are roots of the equation ax² + bx + c = 0 (a ≠ 0)

∴ tan P/2 + tan Q/2 = 

Now consider, 

⇒ a – c = – b ⇒ a + b = c

f(θ) = sinθ (sinθ + sin ³θ)

= (sin θ + 3 sinθ - 4 sin θ). sinθ

= (4 sinθ - 4 sin ³θ) sinθ = sin²θ (4 - 4 sin²θ)

= 4 sin²θ (1– sin²θ)

= 4 sin²θ cos²θ = (2 sinθ cosθ)² = (sin 2θ)² ≥ 0

which is true for all θ.

To simplify the det. Let sin x = a; cos x = b the equation becomes

= 0 Operating C₂ – C₁; C₃ – C₂ we get

⇒ a (a – b)² – (b – a) [b(a – b) – b (b – a)] = 0
⇒ a (a – b)² – 2b (b – a) (a – b) = 0
⇒ (a – b)² (a – 2b) = 0
⇒ (a = b) or a =2b

⇒ tan x = 1 or tan x = 2. But we have -π/ 4 ≤ x ≤π/4

⇒ tan ( -π/4) ≤ tan x ≤ tan (π/4) ⇒ -1 ≤ tan x≤1

∴ tan x = 1 ⇒ x = π/4   ∴ Only one real root is there.

 We are given that (cot α₁). (cot α₂ ) .... (cot α_n) = 1
⇒ (cos α₁) (cos α₂) .... (cos α_n)  
= (sin α₁)  (sin α₂) ....(sin α_n) ....(1)
Let y = (cos α₁) (cos α₂) .... (cos α_n)  (to be max.)
Squaring both sides,
we get y² = (cos² α₁)  ( cos² α₂) .... (cos² α_n)          
= cos α₁ sin α₁ cos α₂ sin α₂ .... cos α_n sin α_n(Using (1))

[sin 2α1 sin 2α2 .... sin2 αn ]

∴ Max. value of y is 1/2^n/2.

Given that α + β = π/2 ⇒ α = π/2 -β

We know that

NOTE THIS STEP

(considering only integral values)
⇒ k can take 8 integral values.

Given that sinθ = 1 /2 and cosφ = 1 /3 and θ and φ both are acute angles

Given that cos (α - β) = 1 and cos (α + β) = 1/e

where  α,β ∈ [-π,π]

Now, cos (α - β) = 1 ⇒ α - β = 0  ⇒ α = β

Now, cos (α + β) = 1/e ⇒ cos 2α = 1/e

⇒ There will be two values of 2α satisfying cos2α = 1/e in [0, 2π ] and two in [– 2π , 0].

⇒ There will be four values of α in [-π,π] and correspondingly four values of β. Hence there are four sets of (α, β).

2 sin²θ - 5 sinθ + 2 > 0

⇒ (sinθ - 2) (2 sinθ - 1) > 0

From graph, we get 

 tanθ < 1 and cotθ >1

Let  tanθ = 1 - x and cotθ = 1+y

Where x, y > 0 and are very small, then

NOTE THIS STEP

2 sin²θ - cos 2θ=0 ⇒1 - 2 cos 2θ = 0

 ....(1)

where q ∈ [0, 2π] Also 2 cos 2θ - 3 sinθ=0 ⇒ 2 sin2θ + 3 sinθ – 2 = 0

⇒ (2 sinθ – 1) (sinθ + 2) = 0

[∵ sinθ ≠-2]

.... (2),    where q ∈ [0, 2π]

Combining (1) and (2), we ge

∴ Two solutions are there.

sin x + 2sin 2x – sin 3x = 3
⇒ sin x + 4sin x cos x – 3sin x + 4sin³ x = 3
⇒ sin x (–2 + 2cos x + 4sin² x) = 3
⇒ sin x (– 2 + 2cos x + 4 – 4cos² x) = 3

⇒ 2 + 2 cos x – 

⇒

⇒ 

LHS  and RHS ≥ 3

∴  Equation has no solution.

sec x + cosec x + 2 (tan x – cot x) = 0

= cos²x – sin² x

For x ∈ S, n = 0

∴ Sum of all values of x  =