JEE Advanced Level Test: Trigonometric Equations- 2 - NDA MCQ

Total number of solutions of sin x . tan 4x = cos x belonging to (0, π) are

Cos²θ/2 =

If 2 tan²x – 5 sec x – 1 = 0 has 7 different roots in , n ∈ N, then greatest value of n is

The values of x between 0 and 2p which satisfy the equation sinx . = 1 are in A.P. The common difference of the A.P. is

Equation of the form P (sinx ± cosx, sinx cosx) = 0 where P(y, z) is a polynomial, can be solved by the change :

cos x ± sin x = t ; 1 ± 2 sin x cos x = t². Reduce the given equation into P = 0

sin x + cos = 1 + sin x cos x is

If (cos x – sin x)  + 2 = 0, then x is

sin⁴x + cos⁴x = sin x cos x then x is–

The minimum value of 27^cos2x 81^sin2x is

Number of roots of the equation cos⁷ x + sin⁴ x = 1 in the interval [0, 2π] is

The smallest positive number of π for whcih the equation cos (π sin x) = sin (π cos x) has a solutionin [0, 2π] is

If sin θ + 7 cos θ = 5, then tan (θ/2) is a root of the equation

Match the following for number of solutions in [0, 2p]

Column - I Column - II

(A) sin² q _ tan² q = 1 (P) 2

(B) sin q + cos q = 1 (Q) 0

(C) tan q + sec q = 2cosq (R) 3

(D) 3sin²q _ 4 sinq + 1 = 0 (S) 1

What are the most general values of q which satisfy the equations,

(a) sin q =  (b) tan (x _ 1) =  (c) tanq = _1 (d) cosec q = 

(e) 2cot²q = cosec²q

If 0 < t < 2π and sin t = - 1, then t =

Solve : cot q + tan q = 2 cosec q

Solve : sin2q = cos 3q

Solve : cot q = tan 8q

Solve : tan²q _ (1 + ) tan q +  = 0

 Find all the angles between 0º and 90º which satisfy the equation sec²q . cosec²q + 2 cosec²q = 8

Solve : 4 cos q _ 3 sec q = 2 tan q

Solve : cot q _ tan q = 2

Solve : sin q + sin 3q + sin 5q = 0