How do you solve these...?

Question

What is the trick to solving them? Please provide the steps and all the answers. Thanks.

sin (60 degrees-x)=2sin x

sinx=sin2x

sinxcosx=.5

Answer 1

sin(60-x)=
sin(60)cos(x)+sin(x)cos(60)
√3/2cos(x)+1/2sin(x)=2sin(x)

√3/2cos(x)=3/2sin(x)

tan(x)=√3/3 ==>x=30

sin(x)=2sin(x)cos(x) ==>
sin(x)-2sin(x)cos(x)=0
sin(x)(1-2cos(x))=0
so sin(x)=0 or cos(x)=1/2
so x=0 or x=60

sin(x)cos(x)=1/2
multiply both sides by 2

2sin(x)cos(x)=1
but this is ==>
sin(2x)=1
2x=90
x=45

Answer 2

sin(60 - x) = 2sin(x)

Use the sine subtraction identity, which goes
sin(a + b) = sin(a)cos(b) - sin(b)cos(a).

sin(60)cos(x) - sin(x)cos(60) = 2sin(x)

We know the sine and cosine of 60 degrees; it's sqrt(3)/2 and 1/2 respectively.

[sqrt(3)/2]cos(x) - sin(x)(1/2) = 2sin(x)
Multiply everything by 2,

sqrt(3) cos(x) - sin(x) = 4sin(x)
Move the sin(x) to one side and the cos(x) to the other side.

sqrt(3) cos(x) = 5sin(x)

Divide both sides by 5cos(x),

sqrt(3)/5 = sin(x)/cos(x)

sqrt(3)/5 = tan(x)

This isn't one of our known values, so the answer within the restriction of -pi/2 to pi/2 is

x = arctan( sqrt(3)/5 )