Solve for all values [0,2pi)

Question

sin (7x) - sin(x) - cos(4x) = 0

Using identities.

(I can solve it using other methods, but using trigonometric identities, nothing simple has come to my mind. I can only represent it in powers of sin(x), which is very lengthy)

Thank you.

Answer 1

sin 7x - sin x - cos4x = 0

sin 7x - sin x = cos 4x

as sin(7x) - sin(x) = sin(4x+3x) - sin(4x - 3x) = sin 4x cos 3x + cos4xsin 3x - (sin 4x cos -3x + cos 4x sin -3x) =

sin 4x cos 3x + cos4xsin 3x - sin 4x cos 3x + cos 4x sin 3x = 2 cos 4x sin 3x, we have

2 cos 4x sin 3x = cos4x

cos4x(2sin3x - 1) = 0

cos4x =0; sin3x = 1/2

cos4x = 0; cos x = 0 at π/2 and 3π/2, so cos 4x = 0 at π/8, 3π/8, 5π/8, 7π/8, 9π/8, 11π/8, 13π/8, 15π/8 (divide by 4 and add π/2, π, 3π/2)

sin 3x = 1/2; sin x = 1/2 at π/6 and 5π/6,so sin 3x = 1/2 at π/18, 5π/18, 13π/18, 17π/18, 25π/18, 29π/18

(divide by 3 and add 2π/3, 4π/3)

If we want the solutions in order, we have π/18, π/8, 5π/18, 3π/8, 5π/8, 13π/18, 7π/8, 17π/18, 9π/8, 11π/8, 25π/18, 29π/18, 13π/8, 15π/8

Answer 2

sinx - sin7x - cos4x = 0

-2cos4xsin3x - cos4x = 0; cos4x(2sin3x + 1) = 0

cos4x = 0, 4x = π/2 + πn; x = π/8 + πn/4; n = 0, 1, 2,...,7

sin3x = - 1/2; 3x = 7π/6 + 2πn, x = 7π/18 + 2πn/3 or 3x = 11π/6 + 2πn, x = 11π/18 + 2πn/3; n = 0, 1, 2

If you have + cos4x then you have sin3x = 1/2 = 1/2 3x = π/6 + πn, x = π/18 + 2πn/3

or 3x = 5π/6 + 2πn, x = 5π/18 + 2πn/3; n = 0, 1, 2