To solve this equation, we can use the trigonometric identity $sin^2x + cos^2x = 1$.

First, let us rewrite $sin^2x$ and $cos^2x$ in terms of $sinx$ and $cosx$:

$sin^2x = 1 - cos^2x$

Substitute this into the equation:

$9sinxcosx - 7cos^2x = 2(1 - cos^2x)$

Expand and simplify:

$9sinxcosx - 7cos^2x = 2 - 2cos^2x$

Rearrange terms:

$9sinxcosx + 2cos^2x - 7cos^2x = 2$

$9sinxcosx - 5cos^2x = 2$

Now we will use the trigonometric identity $sin2x = 2sinxcosx$:

$sin2x = 2sinxcosx$

Substitute $sin2x$ into the equation:

$9/2sin2x - 5cos^2x = 2$

$9/2sin2x - 5(1 - sin^2x) = 2$

$9/2sin2x - 5 + 5sin^2x = 2$

$9/2sin2x + 5sin^2x - 5 = 2$

$9/2sin2x + 5sin^2x = 7$

$sin2x(9/2 + 5) = 7$

$sin2x(19/2) = 7$

$sin2x = 14/(19)$

By using the double angle formula $sin2x = 2sinxcosx$, we can determine sine and cosine values for $x$.