Multiple Choice Questions on Trigonometric Equations Class 11 OP Malhotra Exe-6C ISC Maths Solutions Ch-6. In this article you would learn to solve all type MCQs on Trigonometric Equations. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11

Trigonometric Equations Class 11 OP Malhotra Multiple Choice Questions ISC Maths Solutions Ch-6

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-6	Trigonometric Equations.
Writer	OP Malhotra
Exe-6(C)	Multiple Choice Questions.

Multiple Choice Questions on Trigonometric Equations

OP Malhotra ISC Class 11 Maths Solutions

Que-1: If sin2x = 4cosx, then x is equal to

(a) nπ/2 + π/4, n ∈ Z   
(b) no value   
(c) nπ + (−1)ⁿ π/4, n ∈ Z   
(d) 2nπ ± π/2, n ∈ Z

Sol: (d) 2nπ ± π/2, n ∈ Z
sin2x = 2sinx cosx = 4cosx
⇒ 2sinx = 4 ⇒ sinx = 2 (not possible) OR cosx = 0
⇒ x = 2nπ ± π/2

Que-2: If sinθ = √3 cosθ, −π < θ < 0, then θ is equal to

(a) −5π/6   
(b) −2π/3   
(c) 2π/3   
(d) 5π/6

Sol: (b) −2π/3
sinθ / cosθ = √3 ⇒ tanθ = √3
Reference angle = π/3
Given −π < θ < 0 ⇒ θ = −2π/3

Que-3: If the ∠A lies in the third quadrant and it satisfies the equation 4(sin²x + cosx) = 1, then what is the measure of ∠A?

(a) 225°   
(b) 240°   
(c) 210°   
(d) None of these

Sol: (b) 240°
4(sin²A + cosA) = 1
⇒ 4(1 − cos²A + cosA) = 1
⇒ 4 − 4cos²A + 4cosA = 1
⇒ 4cos²A − 4cosA − 3 = 0
⇒ (2cosA − 3)(2cosA + 1) = 0
⇒ cosA = −1/2 (since A is in 3rd quadrant)
⇒ A = 240°

Que-4: The general solution of tan²θ = 3 is

(a) nπ ± π/3   
(b) nπ + (−1)ⁿ π/3   
(c) 2nπ ± π/3   
(d) 2nπ + (−1)ⁿ π/3

Sol: (a) nπ ± π/3
tan²θ = 3 ⇒ tanθ = ±√3
⇒ θ = nπ ± π/3

Que-5: The general solution of cotθ + tanθ = 2 is

(a) θ = nπ/2 + (−1)ⁿ π/8
(b) θ = nπ/2 + (−1)ⁿ π/4
(c) θ = nπ/2 + (−1)ⁿ π/6
(d) θ = nπ + (−1)ⁿ π/8

Sol: (b) θ = nπ/2 + (−1)ⁿ π/4
cotθ + tanθ = (cos²θ + sin²θ)/(sinθ cosθ) = 1/(sinθ cosθ)
So, 1/(sinθ cosθ) = 2 ⇒ sinθ cosθ = 1/2
⇒ sin2θ = 1 ⇒ 2θ = nπ + π/2
⇒ θ = nπ/2 + (−1)ⁿ π/4

Que-6: The number of solutions of the equation sin²θ − cos2θ = 5/4 (0 ≤ θ ≤ 2π) is

(a) 1       (b) 2        (c) 3        (d) 4

Sol: (d) 4
cos2θ = 1 − 2sin²θ
⇒ sin²θ − (1 − 2sin²θ) = 5/4
⇒ 3sin²θ − 1 = 5/4
⇒ sin²θ = 3/4 ⇒ sinθ = ±√3/2
In [0, 2π], solutions = 4

Que-7: The number of values of n in the interval [0, 3π] satisfying the equation 2sin²x + 5sinx − 3 = 0 is

(a) 1       (b) 2        (c) 4         (d) 6

Sol: (c) 4
2sin²x + 5sinx − 3 = 0
⇒ (2sinx − 1)(sinx + 3) = 0
⇒ sinx = 1/2 (valid), sinx = −3 (invalid)
sinx = 1/2 ⇒ solutions in [0, 3π] = 4

Que-8: If tanθ + tan4θ + tan7θ = tanθ tan4θ tan7θ, then the value of θ is

(a) nπ   
(b) nπ/2   
(c) nπ/6   
(d) nπ/12

Sol: (d) nπ/12
Using identity: tanA + tanB + tanC = tanA tanB tanC ⟹ A + B + C = nπ
θ + 4θ + 7θ = 12θ = nπ
∴ θ = nπ/12

Que-9: The smallest value of θ satisfying the equation √3 (cotθ + tanθ) = 4 is

(a) π/12   
(b) π/6   
(c) π/3   
(d) 2π/3

Sol: (b) π/6
√3 (cotθ + tanθ) = 4
cotθ + tanθ = (cos²θ + sin²θ)/(sinθ cosθ) = 1/(sinθ cosθ)
⇒ √3 / (sinθ cosθ) = 4
⇒ sinθ cosθ = √3 / 4
⇒ (1/2) sin2θ = √3 / 4
⇒ sin2θ = √3 / 2
⇒ 2θ = π/3 ⇒ θ = π/6

Que-10: The general solution of the trigonometrical equation 2 + sec2x − sec4x = 0 is

(a) nπ/2 + π/8
(b) (2n+1)π/10
(c) nπ/4 + π/8
(d) (2n+1)π/12

Sol: (b) (2n+1)π/10
2 + sec2x − sec4x = 0
⇒ sec4x − sec2x = 2
Using identity: sec4x = 1/cos4x and cos4x = 2cos²2x − 1
Solving gives: 10x = (2n+1)π
∴ x = (2n+1)π/10

Que-11: The number of solutions of the equation tanx + secx = 2cosx lying in the interval [0, 2π] is

(a) 0        (b) 1        (c) 2         (d) 3

Sol: (c) 2
Write in sin-cos form:
sinx/cosx + 1/cosx = 2cosx
(sinx + 1)/cosx = 2cosx
sinx + 1 = 2cos²x
Use cos²x = 1 − sin²x → solve → gives two valid solutions in [0,2π]

Que-12: Which one of the following is one of the solutions of the equation tan2θ tanθ = 1?

(a) π/12   
(b) π/6   
(c) π/4   
(d) π/3

Sol: (b) π/6
tan2θ = 2tanθ/(1 − tan²θ)
So equation becomes:
(2tan²θ)/(1 − tan²θ) = 1
Solve → tan²θ = 1/3 → tanθ = 1/√3
θ = π/6

Que-13: The sum of the solutions in (0, 2π) of the equation cosx cos(π/3 − x) cos(π/3 + x) = 1/4 is

(a) π        (b) 2π        (c) 3π        (d) 4π

Sol: (b) 2π
Use identity:
cos(π/3 − x)cos(π/3 + x) = cos²(π/3) − sin²x
= (1/4 − sin²x)
Multiply by cosx and solve → solutions symmetric in (0,2π)

Que-14: The roots of the equation 1 − cosθ = sinθ sin(θ/2) is

(a) kπ/2, k ∈ Z   
(b) kπ, k ∈ Z   
(c) 2kπ, k ∈ Z   
(d) None of these

Sol: (c) 2kπ, k ∈ Z
Use identity:
1 − cosθ = 2sin²(θ/2)
sinθ = 2sin(θ/2)cos(θ/2)
Substitute → simplify → solution θ = 2kπ

Que-15: If tanθ + secθ = √3, then the principal value of θ + π/6 is

(a) π/3       (b) 2π/3        (c) π/4        (d) 3π/4

Sol: (a) π/3
Use identity:
secθ + tanθ = (1 + sinθ)/cosθ
So (1 + sinθ)/cosθ = √3
Solve → θ = π/6
Required: θ + π/6 = π/3

Que-16: The general solution of |sinx| = cosx is given by

(a) nπ − π/4   
(b) nπ + π/4   
(c) nπ ± π/4   
(d) 2nπ ± π/4

Sol: (c) nπ ± π/4
|sinx| = cosx
⇒ sin²x = cos²x
⇒ tan²x = 1
⇒ x = nπ ± π/4

Que-17: The equation √3 sinx + cosx = 4 has

(a) only one solution   
(b) two solutions   
(c) infinitely many solutions   
(d) no solution

Sol: (d) no solution
Maximum value of a sinx + b cosx = √(a²+b²)
= √(3+1)=2 < 4
⇒ impossible

Que-18: If cosx ≠ −1/2, then the solution of cosx + cos2x + cos3x = 0 is

(a) 2nπ ± π/4, n ∈ Z
(b) 2nπ ± π/3, n ∈ Z
(c) 2nπ ± π/6, n ∈ Z
(d) 2nπ ± π/2, n ∈ Z

Sol: (a) 2nπ ± π/4, n ∈ Z
cosx + cos3x = 2cos2x cosx
⇒ equation becomes 2cos2x cosx + cos2x = 0
⇒ cos2x(2cosx+1)=0
Given cosx ≠ −1/2 ⇒ cos2x = 0 ⇒ x = 2nπ ± π/4

Que-19: The set of values of x for which (tan3x − tan2x)/(1 + tan3x tan2x) = 1 is

(a) φ
(b) { π/4 }
(c) { nπ + π/4 , n = 1, 2, 3, … }
(d) { 2nπ + π/4 , n = 1, 2, 3, … }

Sol: (a) φ
LHS = tan(3x − 2x) = tan x
⇒ tanx = 1
⇒ x = nπ + π/4
But denominator undefined ⇒ no valid solution

Que-20: The number of solutions of the equation sinx + sin5x = sin3x lying in the interval [0, π] is

(a) 4          (b) 6        (c) 5         (d) 2

Sol: (b) 6
sinx + sin5x = 2sin3x cos2x = sin3x
⇒ sin 3x (2 cos 2x − 1) = 0
⇒ 6 solutions in [0,π]

Que-21: The sum of the roots of | √3 cosx − sinx | = 2 in [0, 4π] is

(a) 7π/2         (b) 14π/3        (c) 14π        (d) 28π/3

Sol: (d) 28π/3
√3cosx − sinx = 2cos(x+π/6)
⇒ |2cos(x+π/6)| = 2
⇒ |cos(…)| = 1
⇒ roots equally spaced
Total sum in [0,4π] = 28π/3

–: End Trigonometric Equations Class 11 OP Malhotra Exe-6C ISC Maths Ch-6 Solutions :–

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