Question

Find the general solution of
sin x +sin 3x +sin 5x = 0​

Answer 1

Answer:

sinx + sin3x + sin5x = 0

sin3x + (sinx + sin5x) = 0

we know,

sinC + sinD = 2sin(C + D)/2.cos(C-D)/2

sin3x + {2sin(x + 5x)/2.cos(5x-x)/2}= 0

sin3x + {2sin3x.cos2x} = 0

sin3x(1+2cos2x) = 0

Here, sin3x = 0 or cos2x = -1/2

sin3x = 0

3x =nπ

x = nπ/3

again,

cos2x = -1/2

cos2x = -cos(π/3)

cos2x = cos(π -π/3)

cos2x = cos(2π/3)

we know,

If cos∅ = cosA then,

∅ = 2nπ ± A

2x = 2nπ ± (2π/3)

x = nπ ± (π/3)

Hence, the solutions are

x = nπ/3 or nπ ±(π/3)

Answer 2

Answer:

x = n π / 3 OR  x = n π ± 2 π / 3

Step-by-step explanation:

Given :

sin x + sin 3 x + sin 5 x = 0

( sin 5 x + sin x ) + sin 3 x = 0

Using sum to product rule :

sin C + sin D = 2 sin ( ( C + D ) / 2 ) . cos ( ( C - D ) / 2 )

= > 2 sin 3 x . cos 4 x + sin 3 x = 0

Take out sin 3 x common :

= > sin 3 x ( 2 cos 4 x + 1 ) = 0

Case first :

sin 3 x = 0

= >  3 x = n π

= > x = n π / 3

Case second :

2 cos 2 x + 1 = 0

= > 2 cos 2 x = - 1

= > cos 2 x = - 1 / 2

= > cos 2 x = cos ( π - π / 3 )

= > cos 2 x = cos 2 π / 3

We know :

if cos Ф = cos α

= > Ф = 2 n π ± α

= > 2 x = 2 n π ± 2 π / 3

= > x = n π ± 2 π / 3

Therefore we get required answer.