Q. 4 G5.0( 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 x + sin 2x + sin 3x + sin 4x = 0

To solve the equation we need to change its form so that we can equate the t-ratios individually.

For this we will be applying transformation formulae. While applying the

Transformation formula we need to select the terms wisely which we want

to transform.

As, sin x + sin 2x + sin 3x + sin 4x= 0

we will use sin x and sin 3x together in 1 group for transformation and sin 4x and sin 2x common in other group as after transformation both will give cos x term which can be taken common.

{ sin A + sin B = (sin x + sin 3x )+( sin 2x + sin 4x) = 0

2 sin + 2 sin 2sin 2x cos x + 2sin 3x cos x= 0

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

Again using transformation formula, we have:

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

In case of sin x = 0 we can give solution directly as sin x = 0 is true for x = integral multiple of π  where n,p,m ϵ Z

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