We have to solve sin 2t - 3 = 6 sin t - cos t for x.

Now sin 2t = 2 sin t cos t

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

=> 2sin t *cos t -3 = 6 sin t - cos t

=> 2sin t *cos t - 6 sin t + cos t - 3 = 0

=> 2 sin t( cost - 3) +1 ( cos t - 3) = 0

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

For 2 sin t + 2 = 0

=> sin t = -2/2 = -1

=> t = arc sin (-1)= -90 degree

For cos t -3 =0, we don't have any value of t as cos t > 1.

**Therefore t = -90 degrees.**

We'll move all terms of the equation, to the left side:

sin 2t - 3 - 6 sin t + cost = 0

We'll re-write the equation, substituteing sin2t = 2sintcost

2sintcost - 6 sin t + cost - 3 = 0

We could also factorize the first term and the3rd term, by cos t:

cos t(2sin t+1) - 6 sin t - 3 = 0

We'll also factorize the last 2 terms by -3:

cos t(2sin t+1) - 3(2sin t + 1) = 0

We'll factorize again by 2sin t+1:

(2sin t+1)(cos t - 3) = 0

We'll set the first factor as zero:

2sin t + 1 = 0

We'll subtract 1:

2sin t = -1

sin t = -1/2

t = arcsin (-1/2)

Since we have to solve the equation in the range (0,2pi), we'll validate the solution from the unit circle.

The sine function is negative in the 3rd and 4th quadrant, so the solutions for t are:

t = pi + pi/6

**t = 7pi/6 (3rd quadrant)**

t = 2pi - pi/6

**t = 11pi/6 (4th quadrant)**

We'll set the 2nd factor as zero:

cos t - 3 = 0

We'll add 3 both sides:

cost = 3

There are no solutions for t in this case, since the value of the function cosine is not bigger than 1.

**There are only 2 valid solutions for the given equation, in the range (0,2pi): {7pi/6 ; 11pi/6}.**

sin2t-3 = 6sint-cost.

We write sin2t = 2sintcost and rewrite the given equation:

2sint*cost -3 = 6sint - cost.

2sintcost - 6sint - 3 + cost = 0.

2sint(cost - 3)+1(-3 + cost) = 0.

2sint(cost - 3) +1(cost - 3) = 0.

(cost-3)(2sint+1) = 0.

cost-3 = 0 gives cost = 3 which is invalid as |cost| < = 1 always.

2sint +1 = 0 gives 2sint = -1, or sint = -1/2.

So t = 7pi/6, or 11pi/6.