Solve 6 sin ( 2α)− 1 sin ( α ) = 0   for all solutions   0 ≤ α < 2 π.   ?

6 sin(2α) - sin(α) = 0
6 [2 sin(α) cos(α)] - sin(α) = 0
12 sin(α) cos(α) - sin(α) = 0
sin(α) [12 cos(α) - 1] = 0
sin(α) = 0   or   cos(α) = 1/12
α = 0, π      or   α = 0.473, 2π - 0.473
Hence, α = 0, 0.473, π, 5.81

Rule
sin(2 α )= 2sin α cos α

Let A=alpha
6sin(2A)-sin(A)=0, 0=<A<=2pi.
=>
12sin(A)cos(A)-sin(A)=0
=>
sin(A)[12cos(A)-1]=0
=>
sin(A)=0=>A=0, pi, 2pi.
12cos(A)-1=0=>cos(A)=1/12=>A=1.48736624, 4.795819067.
Thus, the solution set is approximated to
{0, 1.487, 3.142, 4.796, 6.283}

Your notation is ambiguous.
Did you mean 6sin(2α) − sin(α) = 0 ?

uhhhh the answer is 2

Let me replace Alpha by x. So Alpha = x

To evaluate x within Range  0 ≤ x < 2 π. 

6 sin (2x) - sin x  =  0

=> 6 sin x cos x  -  sin x  =  0

=>  sin x ( 6 cos x - 1 )  =  0

=>  sin x  =  0  and  cos x  =  1/6

When  sin x  =  0 ,  x  =  0,  π    Radians in the given Range. 

When  cos x  =  1/6 ,  x  =  1.40 Radian, 4.88 Radian

x ( Or Alpha )  =  0, π , 1.40 Radian, 4.88 Radian  .......... Answer

6sin(2α) - sin(α) = 0
12sin(α)cos(α) - sinα = 0
sin(α)[12cos(α) - 1] = 0

0 ≤ α < 2π
sin(α) = 0 → α = 0, π
12cos(α) - 1 = 0 → α = arccos(1/2), 2π - arccos(1/12)
α = 1.48737, 4.79582

Ans:
α = 0, 1.48737, π, 4.79582

6 sin (2α) - 1 sin (α) = 0
6 (2 sin α cos α) - sin α - 0
sin α (12 cos α - 1) - 0

          ⇒ Either sin α = 0 or 12 cos α -1 = 0

          case 1. sin α = 0    ⇒  α = 0, π, 2π  ..................ANS

          case 2. 12 cos α - 1 = 0  
                                        ⇒  cos α = 1/12
                                             α = cos⁻¹ (1/12)
                                             α = 0.47π  .......................ANS