sin^2 y = 5siny - 6

We will move all terms to the left side so the right side is 0.

==> sin^2 y - 5siny + 6 = 0

Now let us assume that u= siny

==> u^2 - 5u + 6 = 0

Let us factor.

==> (u-2)(u-3) = 0

==> u= 2 ==> siny = 2

==> u= 3 ==> siny = 3

Both solutions are invalid because the the values of sine must be between -1 and 1.

==> -1 =< siny =< 1

**Then, the function has no solution.**

The question states that (sin y)^2 = 5* sin y - 6.

let x = sin y

=> x^2 = 5x - 6

=> x^2 - 5x + 6 = 0

=> x^2 - 3x - 2x + 6 = 0

=> x(x -3) - 2( x - 3) =0

=> (x - 2) (x -3) = 0

x has the values 2 and 3.

But as x = sin y, it cannot be greater than 1 or less than -1.

**Therefore the equation has no solutions.**

We'll solve this equation applying the substitution technique. We'll substitute sin y = t and we'll re-write the equation:

t^2 = 5t - 6

We'll subtract 5t - 6 both sides:

t^2 - 5t + 6 = 0

We'll apply quadratic formula:

t1 = [5 + sqrt(25 - 24)]/2

t1 = (5 + 1)/2

t1 = 3

t2 = (5 - 1)/2

t2 = 2

But sin y = t

We'll put sin y = t1

**sin y = 3 impossible!**

sin y = t2

**sin y = 2 impossible!**

**The equation has no valid solutions!**

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