Q. 4 D5.0( 1 Vote )

# Solve the followi

Answer :

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ny, where n Z.

• cos x = cos y, implies x = 2nπ ± y, where n Z.

• tan x = tan y, implies x = nπ + y, where n Z.

Given,

cos x cos 2x cos 3x = �

4cos x cos 2x cos 3x – 1 = 0

{ 2 cos A cos B = cos (A + B) + cos (A – B)}

2(2cos x cos 3x)cos 2x – 1 = 0

2(cos 4x + cos 2x)cos2x – 1 = 0

2(2cos2 2x – 1 + cos 2x)cos 2x – 1 = 0 {using cos 2θ = 2cos2θ – 1 }

4cos3 2x – 2cos 2x + 2cos2 2x – 1 = 0

2cos2 2x (2cos 2x + 1) -1(2cos 2x + 1) = 0

(2cos2 2x – 1)(2 cos 2x + 1) = 0

either, 2cos 2x + 1 = 0 or (2cos2 2x – 1) = 0

cos 2x = -1/2 or cos 4x = 0 {using cos 2θ = 2cos2θ – 1}

cos 2x = cos (π - π/3) = cos 2π /3 or cos 4x = cos π/2

If cos x = cos y implies x = 2nπ ± y, where n Z.

In case of cos x = 0 we can give solution directly as cos x = 0 is true for x = odd multiple of π/2

Comparing obtained equation with standard equation, we have:

y = 2π / 3 or y = π/2

2x = 2mπ ± 2π/3 or 4x = (2n+1)π/2 where m,n ϵ Z ….ans

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses RELATED QUESTIONS :

Solve the followiRD Sharma - Mathematics

3sin2 RD Sharma - Mathematics

Solve the followiRD Sharma - Mathematics

Solve : <span lanRD Sharma - Mathematics

Solve the followiRD Sharma - Mathematics

Solve : <span lanRD Sharma - Mathematics

Solve the followiRD Sharma - Mathematics

Solve the followiRD Sharma - Mathematics

Find the general RD Sharma - Mathematics

Find the general RD Sharma - Mathematics