If sin 3 theta - sin theta = 4 cos ^(2)...

If ` sin 3 theta - sin theta = 4 cos ^(2) theta - 2 " then " theta = `

A

`{(2 n + 1) (pi)/( 4), n in Z} cup {(2 n + 1) (pi)/( 2) , n in Z}`

B

`{(2 n + 1) (pi)/( 2), n in Z} cup {(4 n + 1) (pi)/( 2) , n in Z}`

C

`{(2 n + 1) (pi)/( 4), n in Z} cup {(4 n + 1) (pi)/( 2) , n in Z}`

D

`{(4 n + 1) (pi)/( 4), n in Z} cup {(4 n + 1) (pi)/( 2) , n in Z}`

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The correct Answer is:

C

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