To solve the equation sin²x - cos²x - 3sinx + 2 = 0, we can make use of the trigonometric identity sin²x + cos²x = 1.

First, we can rewrite the equation by substituting sin²x + cos²x for 1:

1 - 2cos²x - 3sinx + 2 = 0

Simplifying this equation, we get:

-2cos²x - 3sinx + 3 = 0

Now, we can use the Pythagorean identity cos²x = 1 - sin²x to rewrite the equation in terms of sinx:

-2$1-si{n}^{2}x$ - 3sinx + 3 = 0

Expanding and simplifying:

-2 + 2sin²x - 3sinx + 3 = 0
2sin²x - 3sinx + 1 = 0

Now, we have a quadratic equation in terms of sinx. We can solve this by factoring or using the quadratic formula. The equation factors to:

$2sinx-1$$sinx-1$ = 0

Setting each factor to zero and solving for sinx, we get:

2sinx - 1 = 0
sinx = 1/2

sinx - 1 = 0
sinx = 1

Therefore, the solutions to the original equation sin²x - cos²x - 3sinx + 2 = 0 are sinx = 1/2 and sinx = 1.