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,


sin 3x – sin x = 4 cos2 x – 2


sin 3x – sin x = 2(2 cos2 x – 1)


sin 3x – sin x = 2 cos 2x { cos 2θ = 2cos2 θ – 1}


{ sin A - sin B =





either, cos 2x = 0 or sin x = 1 = sin π/2


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


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



where m, n ϵ Z


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
caricature
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 RS Aggarwal - Mathematics