Solve for t in (0,2pi) sin 2t - 3 = 6 sin t - cost

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}.