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# Advanced Math/Pre-cal trigonometric question

Question
Solve algebraically for x in the domain [0, 2 pi]
sin(2x)=sin(x/2)

If sin(2x) = sin(x/2), then

1-sinē(2x) = 1-sinē(x/2) and

cosē(2x) = cosē(x/2)

So cos(2x)=cos(x/2) or cos(2x)=-cos(x/2)

If cos(2x)=cos(x/2) ,

Sin(3x/2) = sin( 2x-x/2 ) = sin(2x)cos(x/2) - sin(x/2)cos(2x) = 0

So 3x/2  = nπ for some integer n

Then x = 2nπ/3

Recalling that 0 ≤ x ≤  2π

we see that 0 , 2π/3  , 4π/3 ,  2π  are the only possible values for x in this case.

Checking each of these , we find that 0 , 4π/3 ,  2π  are the only values for x which satisfy sin(2x)=sin(x/2) .

Next , suppose  cos(2x) = -cos(x/2)

Sin(5x/2) = sin( 2x+x/2 ) = sin(2x)cos(x/2) + sin(x/2)cos(2x) = 0

So 5x/2  = nπ for some integer n

Then x = 2nπ/5

Recalling that 0 ≤ x ≤  2π

we see that 0 , 2π/5  , 4π/5 ,  6π/5 ,  8π/5 ,  2π  are the only possible values for x in this case.

Checking each of these , we find  0,  2π/5 ,  6π/5 ,  2π are the only values for x which satisfy sin(2x)=sin(x/2) .

Combining the above results , the solutions to the equation are

{0,  2π/5 ,  6π/5 , 4π/3 ,  2π }
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#### Socrates

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