To solve the equation 2sin^2x + 3sinx × cosx - 5cos^2x = 0, we can use the fact that sin^2x + cos^2x = 1.

First, let's rewrite the equation using this identity:

2(1 - cos^2x) + 3sinx × cosx - 5cos^2x = 0

Expanding and rearranging the terms:

2 - 2cos^2x + 3sinx × cosx - 5cos^2x = 0

Combining like terms:

-7cos^2x + 3sinx × cosx + 2 = 0

Now, let's factor out cosx:

cosx(-7cosx + 3sinx) + 2 = 0

Since cosx cannot be zero (as it would make the equation undefined), we can divide both sides of the equation by cosx:

-7cosx + 3sinx + 2/cosx = 0

Simplifying:

-7 + 3tanx + 2secx = 0

Now, we have a trigonometric equation involving tangent (tanx) and secant (secx). To solve this equation, we can substitute secx = 1/cosx and tanx = sinx/cosx:

-7 + 3(sinx/cosx) + 2(1/cosx) = 0

Multiplying through by cosx:

-7cosx + 3sinx + 2 = 0

This is a linear equation in terms of sinx and cosx. We can solve it using standard algebraic methods.