You need to use the double angle identity, such that:

`sin 2t = 2 sin t cos t`

You need to write the equation in terms of `sin t` and `cos t` , such that:

`2 sin t cos t - 3 = 6 sin t - cos t`

You need to group the terms such that:

`2 sin t cos t - 6 sin t = 3 - cos t`

Factoring out 2sin (t) to the left side yields:

`2sin (t) (cos (t) - 3 ) = -(cos (t) - 3) => 2sin (t) (cos (t) - 3) + (cos (t) - 3) = 0`

You need to factor out (`cos t - 3)` such that:

`(cos t - 3)(2sin t + 1) = 0 => {(cos t - 3 = 0),(2 sin t + 1 = 0):}`

`cos t = 3` invalid , `cos t in [-1,1]`

`2 sin (t) = -1 => sin (t) = -1/2`

Sine is negative for `t in (pi,(3pi)/2) U ((3pi)/2, 2pi)` , such that:

`t = pi + pi/6 => t = (7pi)/6`

`t = 2pi - pi/6 => t = (11pi)/6`

**Hence, evaluating the solutions to the given equation, under the given conditions, yields **`t = (7pi)/6 ; t = (11pi)/6.`

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