RD Sharma Solutions for Class 11 Maths Chapter 11: Trigonometric Equations

RD Sharma Solutions for Class 11 Maths Chapter 11: Trigonometric Equations are available here to help students effectively prepare for their CBSE Class 11 Syllabus and final mathematics tests. To excel in Class 11 trigonometry, it is crucial for students to thoroughly practice every question from all exercises in this chapter. This chapter covers important trigonometric formulas, general solutions of trigonometric equations, principal solutions, and various methods to solve trigonometric equations.

The RD Sharma Solution for Class 11 have been carefully designed by experts at Infinity Learn as a reliable study resource and reference guide to help students clear their doubts quickly and achieve high marks in exams. Whether you're preparing for CBSE Class 11 Maths or other state boards, these step-by-step solutions ensure a deep understanding of key concepts and formulas essential for success in trigonometry and overall mathematics.

RD Sharma Solutions Class 11 Maths Chapter 11 – Free PDF Download

RD Sharma Class 11 Solutions Chapter 11 – Trigonometric Equations comprises detailed exercises designed to cover the entire syllabus prescribed by CBSE for Class 11 Mathematics. These expertly curated solutions facilitate easy and quick calculations, helping students to master important concepts efficiently. Students can download the RD Sharma Solutions PDF for Class 11 Maths Chapter 11 Trigonometric Equations from the links provided below for convenient offline study.

This chapter focuses on several key trigonometric topics, including:

• The General Solutions of Trigonometric Equations
• Solving Trigonometric Equations of the form sinx=siny, cosx=cosy, and tanx=tany
• Solving Trigonometric Equations by Factorisation
• Solving Trigonometric Equations Reducible to Quadratic Equations
• Solving Trigonometric Equations by Auxiliary Angle Method (acosx+bsinx=c)
• Solving Trigonometric Equations involving Multiple Angles

These topics form the foundation for solving complex problems in trigonometry and have practical applications in fields such as engineering and physics.

RD Sharma Solutions for Class 11 Maths Trigonometric Equations: Step by Step Solutions

RD Sharma Class 11 Maths Chapter 11: Trigonometric Equations

Question 1: Find the general solution of: sin x = √3/2

Step 1: Identify the angle whose sine is √3/2

We know that sin 60° = sin π/3 = √3/2

Step 2: Write the general solution

For sin x = sin α, the general solution is:

x = nπ + (-1)ⁿα, where n ∈ Z

Step 3: Substitute α = π/3

Answer: x = nπ + (-1)ⁿ(π/3), where n ∈ Z

Question 2: Solve cos 2θ = -1/2 in the interval [0, 2π]

Step 1: Identify the angle whose cosine is -1/2

We know that cos 120° = cos 2π/3 = -1/2 and cos 240° = cos 4π/3 = -1/2

Step 2: Set up the equation

2θ = 2π/3 or 2θ = 4π/3

Step 3: Solve for θ

θ = π/3 or θ = 2π/3

Step 4: Check if solutions are in [0, 2π]

Answer: θ = π/3, 2π/3

Question 3: Determine all solutions for tan x = 1

Step 1: Identify the angle whose tangent is 1

We know that tan 45° = tan π/4 = 1

Step 2: Write the general solution for tangent

For tan x = tan α, the general solution is:

x = nπ + α, where n ∈ Z

Step 3: Substitute α = π/4

Answer: x = nπ + π/4, where n ∈ Z

Question 4: Write the general solution for sec x = 2

Step 1: Convert to cosine

sec x = 2 means cos x = 1/2

Step 2: Identify the angle whose cosine is 1/2

We know that cos 60° = cos π/3 = 1/2

Step 3: Write the general solution for cosine

For cos x = cos α, the general solution is:

x = 2nπ ± α, where n ∈ Z

Step 4: Substitute α = π/3

Answer: x = 2nπ ± π/3, where n ∈ Z

Question 5: Solve cot 3x = √3

Step 1: Identify the angle whose cotangent is √3

We know that cot 30° = cot π/6 = √3

Step 2: Set up the equation

3x = nπ + π/6, where n ∈ Z

Step 3: Solve for x

x = nπ/3 + π/18, where n ∈ Z

Answer: x = nπ/3 + π/18, where n ∈ Z

Question 6: Find real solutions for 2cos²x - 3cos x + 1 = 0

Step 1: Let y = cos x

The equation becomes: 2y² - 3y + 1 = 0

Step 2: Factor the quadratic

(2y - 1)(y - 1) = 0

Step 3: Solve for y

y = 1/2 or y = 1

Step 4: Substitute back

cos x = 1/2 or cos x = 1

Step 5: Find x values

For cos x = 1/2: x = 2nπ ± π/3

For cos x = 1: x = 2nπ

Answer: x = 2nπ, 2nπ ± π/3, where n ∈ Z

Question 7: Solve 4sin²x + 8cos x - 7 = 0

Step 1: Use the identity sin²x = 1 - cos²x

4(1 - cos²x) + 8cos x - 7 = 0

Step 2: Simplify

4 - 4cos²x + 8cos x - 7 = 0

-4cos²x + 8cos x - 3 = 0

4cos²x - 8cos x + 3 = 0

Step 3: Let y = cos x

4y² - 8y + 3 = 0

Step 4: Use quadratic formula

y = (8 ± √(64 - 48))/8 = (8 ± 4)/8

y = 3/2 or y = 1/2

Step 5: Check validity

cos x = 3/2 is impossible (|cos x| ≤ 1)

cos x = 1/2 is valid

Answer: x = 2nπ ± π/3, where n ∈ Z

Question 8: Determine solutions for tan²x - 2tan x - 3 = 0 in [0, π]

Step 1: Let y = tan x

y² - 2y - 3 = 0

Step 2: Factor

(y - 3)(y + 1) = 0

Step 3: Solve for y

y = 3 or y = -1

Step 4: Substitute back

tan x = 3 or tan x = -1

Step 5: Find x in [0, π]

For tan x = 3: x = arctan(3) ≈ 1.249 radians

For tan x = -1: x = 3π/4

Answer: x = arctan(3), 3π/4

Question 9: Solve √3 cos x + sin x = √2

Step 1: Express in the form R sin(x + α)

√3 cos x + sin x = R sin(x + α)

Step 2: Find R and α

R = √((√3)² + 1²) = √4 = 2

tan α = √3/1 = √3, so α = π/3

Step 3: Rewrite the equation

2 sin(x + π/3) = √2

sin(x + π/3) = √2/2 = 1/√2

Step 4: Solve

x + π/3 = π/4 + 2nπ or x + π/3 = 3π/4 + 2nπ

x = π/4 - π/3 + 2nπ or x = 3π/4 - π/3 + 2nπ

x = -π/12 + 2nπ or x = 5π/12 + 2nπ

Answer: x = -π/12 + 2nπ, 5π/12 + 2nπ, where n ∈ Z

Question 10: Find general solutions for sin x + sin 2x + sin 3x = 0

Step 1: Group terms

(sin x + sin 3x) + sin 2x = 0

Step 2: Use sum-to-product formula

sin A + sin B = 2 sin((A+B)/2) cos((A-B)/2)

sin x + sin 3x = 2 sin(2x) cos(x)

Step 3: Substitute

2 sin(2x) cos(x) + sin 2x = 0

sin 2x (2 cos x + 1) = 0

Step 4: Solve each factor

sin 2x = 0 or 2 cos x + 1 = 0

2x = nπ or cos x = -1/2

x = nπ/2 or x = 2nπ ± 2π/3

Answer: x = nπ/2, 2nπ ± 2π/3, where n ∈ Z

Question 11: Solve sin 3x = cos 2x

Step 1: Use the identity cos θ = sin(π/2 - θ)

sin 3x = sin(π/2 - 2x)

Step 2: Apply general solution for sin A = sin B

3x = nπ + (-1)ⁿ(π/2 - 2x)

Step 3: Consider two cases

Case 1 (n even): 3x = π/2 - 2x + 2kπ

5x = π/2 + 2kπ

x = π/10 + 2kπ/5

Case 2 (n odd): 3x = -(π/2 - 2x) + (2k+1)π

3x = -π/2 + 2x + (2k+1)π

x = π/2 + (2k+1)π

Answer: x = π/10 + 2kπ/5, π/2 + (2k+1)π, where k ∈ Z

Question 12: Find solutions for cos 4x = sin 3x in [0, π]

Step 1: Use cos θ = sin(π/2 - θ)

sin(π/2 - 4x) = sin 3x

Step 2: Apply general solution

π/2 - 4x = nπ + (-1)ⁿ(3x)

Step 3: Consider cases and find solutions in [0, π]

Case 1: π/2 - 4x = 3x + 2kπ

π/2 - 2kπ = 7x

x = (π - 4kπ)/14

Case 2: π/2 - 4x = -3x + (2k+1)π

π/2 - (2k+1)π = x

x = π/2 - (2k+1)π

Step 4: Check which values lie in [0, π]

Answer: x = π/14, 3π/14, 5π/14, 9π/14, 11π/14, 13π/14

Question 13: Determine general solutions for tan 2x = cot(x + π/6)

Step 1: Use cot θ = tan(π/2 - θ)

tan 2x = tan(π/2 - x - π/6)

tan 2x = tan(π/3 - x)

Step 2: Apply general solution for tan A = tan B

2x = nπ + π/3 - x

Step 3: Solve for x

3x = nπ + π/3

x = nπ/3 + π/9

Answer: x = nπ/3 + π/9, where n ∈ Z

Question 14: Solve sin x + sin 5x = sin 3x

Step 1: Use sum-to-product formula

sin x + sin 5x = 2 sin(3x) cos(2x)

Step 2: Substitute into equation

2 sin(3x) cos(2x) = sin 3x

Step 3: Factor

sin 3x (2 cos 2x - 1) = 0

Step 4: Solve each factor

sin 3x = 0 or 2 cos 2x - 1 = 0

3x = nπ or cos 2x = 1/2

x = nπ/3 or 2x = ±π/3 + 2kπ

x = nπ/3 or x = ±π/6 + kπ

Answer: x = nπ/3, ±π/6 + kπ, where n, k ∈ Z

Question 15: Find all x satisfying cos 7x = sin 4x

Step 1: Use cos θ = sin(π/2 - θ)

sin(π/2 - 7x) = sin 4x

Step 2: Apply general solution

π/2 - 7x = nπ + (-1)ⁿ(4x)

Step 3: Consider two cases

Case 1 (n even): π/2 - 7x = 4x + 2kπ

π/2 - 2kπ = 11x

x = (π - 4kπ)/22

Case 2 (n odd): π/2 - 7x = -4x + (2k+1)π

π/2 - (2k+1)π = 3x

x = (π - 2(2k+1)π)/6 = (π - 4kπ - 2π)/6 = -(π + 4kπ)/6

Answer: x = (π - 4kπ)/22, -(π + 4kπ)/6, where k ∈ Z

Question 16: Prove that 2cos²x + 3sin x = 0 has no real solution

Step 1: Use the identity cos²x = 1 - sin²x

2(1 - sin²x) + 3sin x = 0

Step 2: Simplify

2 - 2sin²x + 3sin x = 0

2sin²x - 3sin x - 2 = 0

Step 3: Let y = sin x

2y² - 3y - 2 = 0

Step 4: Use quadratic formula

y = (3 ± √(9 + 16))/4 = (3 ± 5)/4

y = 2 or y = -1/2

Step 5: Check validity

sin x = 2 is impossible (|sin x| ≤ 1)

sin x = -1/2 is valid and gives x = 7π/6 + 2nπ or x = 11π/6 + 2nπ

Correction: The equation actually has solutions. There might be an error in the original problem statement.

Question 17: Find the number of solutions to sin x = x/10 in [-2π, 2π]

Step 1: Analyze the functions

f(x) = sin x (oscillates between -1 and 1)

g(x) = x/10 (linear function)

Step 2: Find intersection points

At x = 0: sin(0) = 0, 0/10 = 0 ✓

Step 3: Check slopes

f'(0) = cos(0) = 1

g'(x) = 1/10 = 0.1

Step 4: Analyze graphically

Since |sin x| ≤ 1 and |x/10| ≤ 2π/10 ≈ 0.628 in [-2π, 2π]

The linear function intersects the sine curve at the origin and two other points (one positive, one negative)

Answer: 3 solutions

Question 18: Solve √(1 + sin 2x) = sin x + cos x

Step 1: Square both sides

1 + sin 2x = (sin x + cos x)²

Step 2: Expand the right side

1 + sin 2x = sin²x + 2sin x cos x + cos²x

Step 3: Use identities

sin²x + cos²x = 1 and 2sin x cos x = sin 2x

1 + sin 2x = 1 + sin 2x

Step 4: Check for restrictions

For the square root to be defined: 1 + sin 2x ≥ 0

This gives sin 2x ≥ -1, which is always true

For the equation to be valid: sin x + cos x ≥ 0

Step 5: Find when sin x + cos x ≥ 0

sin x + cos x = √2 sin(x + π/4) ≥ 0

This occurs when sin(x + π/4) ≥ 0

x + π/4 ∈ [2nπ, (2n+1)π]

x ∈ [2nπ - π/4, (2n+1)π - π/4]

Answer: x ∈ [2nπ - π/4, (2n+1)π - π/4], where n ∈ Z

Question 19: Determine values of k for which cos θ + √3 sin θ = k has solutions

Step 1: Express in R sin(θ + α) form

cos θ + √3 sin θ = R sin(θ + α)

Step 2: Find R

R = √(1² + (√3)²) = √(1 + 3) = 2

Step 3: Find α

tan α = 1/√3, so α = π/6

Step 4: Rewrite equation

2 sin(θ + π/6) = k

sin(θ + π/6) = k/2

Step 5: Apply range of sine function

For solutions to exist: -1 ≤ k/2 ≤ 1

Therefore: -2 ≤ k ≤ 2

Answer: k ∈ [-2, 2]

Question 20: Show that sin⁻¹x = 2cos⁻¹√(1-x²) reduces to 4x⁴ - 4x² + 1 = 0

Step 1: Let α = sin⁻¹x and β = cos⁻¹√(1-x²)

Then sin α = x and cos β = √(1-x²)

Step 2: Use the given equation

α = 2β

sin α = sin(2β)

Step 3: Apply double angle formula

sin(2β) = 2sin β cos β

Step 4: Find sin β

sin β = √(1 - cos²β) = √(1 - (1-x²)) = √x² = |x|

Assuming x ≥ 0, sin β = x

Step 5: Substitute

x = 2x√(1-x²)

Step 6: Square both sides

x² = 4x²(1-x²)

x² = 4x² - 4x⁴

4x⁴ - 3x² = 0

x²(4x² - 3) = 0

Note: This doesn't directly give 4x⁴ - 4x² + 1 = 0. There may be an error in the problem statement or additional steps needed.

Question 21: Solve |sin x| + |cos x| = 0 (where |·| denotes absolute value, not floor function)

Step 1: Analyze the equation

Since |sin x| ≥ 0 and |cos x| ≥ 0 for all real x

Step 2: For the sum to equal zero

Both |sin x| = 0 and |cos x| = 0 must be true simultaneously

Step 3: Check if this is possible

sin x = 0 when x = nπ

cos x = 0 when x = π/2 + nπ

Step 4: Find intersection

There is no value of x for which both sin x = 0 and cos x = 0

This is because sin²x + cos²x = 1 always

Answer: No solution exists

Question 22: Find general solution for sin⁴x + cos⁴x = 5/8

Step 1: Use the identity (a² + b²)² = a⁴ + 2a²b² + b⁴

(sin²x + cos²x)² = sin⁴x + 2sin²x cos²x + cos⁴x

Step 2: Since sin²x + cos²x = 1

1 = sin⁴x + 2sin²x cos²x + cos⁴x

sin⁴x + cos⁴x = 1 - 2sin²x cos²x

Step 3: Substitute into the given equation

1 - 2sin²x cos²x = 5/8

2sin²x cos²x = 1 - 5/8 = 3/8

sin²x cos²x = 3/16

Step 4: Use sin 2x = 2sin x cos x

(sin x cos x)² = 3/16

(sin 2x/2)² = 3/16

sin²(2x) = 3/4

Step 5: Solve for sin 2x

sin 2x = ±√3/2

Step 6: Find general solution

2x = π/3 + nπ or 2x = 2π/3 + nπ or 2x = 4π/3 + nπ or 2x = 5π/3 + nπ

x = π/6 + nπ/2 or x = π/3 + nπ/2

Answer: x = π/6 + nπ/2, π/3 + nπ/2, where n ∈ Z