Q. 4 H5.0( 1 Vote )

# Solve the followi

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

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