Solve trigonometric equation 2sin^2x - 3sinx = - 1.

The first step is to move all terms to the left
side:

2(sin x)^2- 3sinx + 1 =
0

Now, we'll use substitution technique to solve the
equation.

We'll note sin x = t and we'll re-write the
equation in t:

2t^2 - 3t + 1 =
0

Since it is a quadratic, we'll apply the quadratic
formula:

t1 = {-(-3) + sqrt[(-3)^2 -
4*2*1]}/2*2

t1 =
[3+sqrt(9-8)]/4

t1 =
(3+1)/4

t1 = 1

t2 =
(3-1)/4

t2 = 1/2

Now, we'll
put sin x = t1.

sin x =
1

Since it is an elementary equation, we'll apply the
formula:

sin x = a

x = (-1)^k*
arcsin a + 2k*pi

In our case, a =
1:

x =  (-1)^k* arcsin 1 +
2k*pi

x = (-1)^k*(pi/2) +
2k*pi

x = pi/2

Now, we'll
put sin x = t2

sin x = 1/2

x =
(-1)^k* arcsin 1/2 + 2k*pi

x = (-1)^k* (pi/6) +
2k*pi

x = pi/6

x = pi -
pi/6

x =
5pi/6

The solutions of the equation are:{
pi/6 ; pi/2 ; 5pi/6 }.