Bijoy Deka
Last Activity: 8 Years ago
1 + sin 2x = ( sin 3x – cos 3x )2
1 + sin 2x = sin2 3x + cos2 3x – 2sin3x. cos3x
1 + sin 2x = 1 – 2sin3x. cos3x (Since sin2 A + cos2 A = 1)
sin 2x = – sin 6x ( sin 2A = 2sinAcosA)
sin 6x = – sin 2x
sin 6x = sin (π + 2x) ( Since sin (π + A) = – sin A)
6x = nπ – (π + 2x) or 6x = 2nπ + (π + 2x) n ∈ Z
6x + 2x = (n-1)π or 6x -2x = (2n+1)π n ∈ Z
8x = (n-1)π or 4x = (2n +1)π n ∈ Z
x = (n-1)π / 8 or x = (2n +1)π /4 n ∈ Z