Question

(a)
Solve: sin7x + sin 4x + sinx = 0 and 0 < x < 1/2​

Answer 1

Answer:

x = π / 4 or π / 9 or 2π/ 9

Step-by-step explanation:

Given-----> Solve ,

Sin 7x + Sin4x + Sinx = 0 , if , 0 < x < π / 2

Solution----> ATQ,

0 < x < π/2 , so ,

0 < 4x < 2π and 0 < 3x < 3π/2

Sin7x + Sin4x + Sinx = 0

=> Sin7x + Sinx + Sin4x = 0

We know that,

SinC + SinD = 2 Sin ( C + D ) / 2 Cos ( C - D ) / 2

Using it , we get,

=> 2 Sin ( 7x + x )/2 Cos ( 7x - x )/2 + Sin4x = 0

=> 2 Sin ( 8x / 2 ) Cos ( 6x / 2 ) + Sin4x = 0

=> 2 Sin 4x Cos3x + Sin4x = 0

=> Sin4x ( 2 Cos3x + 1 ) = 0

If , Sin4x = 0 , and 0 < 4x < 2π

=> 4x = π

=> x = π / 4

If , 2 Cos3x + 1 = 0

=> 2 Cos3x = - 1

=> Cos 3x = - 1 / 2 , and , 0 < 3x < 3π / 2

=> 3x = 2π/3 , 4π / 3

=> x = 2π/9 , 4π/9