Sinx-3sin2x+sin3x=cosx-3cos2x+cos3x
Answers
Answer:
x = π/8 + n*π/2 (in radians) (OR) x = 22.5° + n * 90° (in degrees)
Step-by-step explanation:
sin x - 3 sin 2x + sin 3x = cos x - 3 cos 2x + cos 3x
sin (2x-x) - 3 sin 2x + sin (2x + x) = cos (2x-x) - 3 cos 2x + cos (2x+x)
sin(2x)cos(x) - sin(x)cos(2x) - 3 sin(2x) + sin(2x)cos(x)+sin(x)cos(2x)
= cos(2x)cos(x) + sin(2x)sin(x) – 3cos(2x) + cos(2x)cos(x) - sin(2x)sin(x)
2 sin(2x)cos(x) - 3 sin(2x) = 2 cos(2x)cos(x) - 3 cos(2x)
sin(2x) * (2 cos(x) - 3) = cos(2x) * (2 cos(x) - 3)
sin(2x) * (2 cos(x) - 3) - cos(2x) * (2 cos(x) - 3) = 0
(2 cos(x) - 3) (sin(2x) - cos(2x)) = 0
//Use zero factor property to solve for x individually:
i.e. 2 cos(x) - 3 = 0 sin(2x) - cos(2x) = 0
2 cosx - 3 = 0 sin 2x = cos 2
cos x = 3/2 Tan 2x = 1
==> No solution here 2x = Tan ⁻¹ (1)
2x = π/4 + n*π where n is any integer
x = π/8 + n*π/2
i.e. in degrees x = 22.5° + n * 90°