What is the general solution for 120sin(x-120)=120sin(x-240) ?

The general solution for 120sin(x-120)=120sin(x-240) is x=90+360n,x=270+360n

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120sin(x-120)=120sin(x-240)

120 sin (x - 12 0^{\circ}) = 120 sin (x - 24 0^{\circ})

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

120 sin (x - 12 0^{\circ}) = 120 sin (x - 24 0^{\circ})

Rewrite using trig identities

120 (- \frac{1}{2} sin (x) - \frac{3}{2} cos (x)) = 120 (- \frac{1}{2} sin (x) + \frac{3}{2} cos (x))

Subtract

120 (- \frac{1}{2} sin (x) + \frac{3}{2} cos (x))

from both sides

- 1203 cos (x) = 0

Divide both sides by

- 1203

cos (x) = 0

General solutions for

cos (x) = 0

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

sin^{2} (x) = 1 - sin^{2} (x)

- 8 cos (θ) = 43

2 sin (3 a) = - 1

sin (b) = 12 (\frac{sin ( 4 2 ^{\circ} )}{22})

arctan (\frac{3}{1}) = tan (θ)

What is the general solution for 120sin(x-120)=120sin(x-240) ?
The general solution for 120sin(x-120)=120sin(x-240) is x=90+360n,x=270+360n