Trigonometric Equation Simplification $$3\sin x + 4\cos x = 2$$
To solve an equation like the one above, we were taught to use the double angle identity formula to get two equations in the form of $R\cos\alpha = y$ where $R$ is a coefficient and $\alpha$ is the second angle being added to $x$ when using the double angle identity. 

Why can't we use the identity $\sin(x) = \cos(x-90)$ to get $3\cos(x-90) + 4\cos x = 2$? Is this equation difficult to simplify further?

Additionally, why was the relationship between $sinx$ and $cosx$ in the pythagorean theorem (modified for the unit circle) not put to use? I did the following:
$$\sin^2x = 1 - \cos^2x$$
$$\therefore \sin x =  ±\sqrt{1-\cos^2x}$$
If:
$$\sin x = y$$
Then,
$$y = ±\sqrt{1-\cos^2x}$$
Meaning that,
$$\cos x = ±\sqrt{1-y^2}$$
Inputting this into the original equation, 
$$3y + 4\sqrt{1-y^2} = 2$$
We see, $$3y-2=-4\sqrt{1-y^2}$$
So, $$(3y-2)^2=16-16y^2$$
Therefore, $$9y^2-12y+4=16-16y^2$$
Rearranging gives, $$25y^2-12y-12=0$$
And so the solutions are, $$y_0,y_1=\frac{12}{50}\pm\frac{1}{50}\sqrt{144+1200}$$
And simplifying yields, $$y_0,y_1=\frac{6\pm4\sqrt{21}}{25}$$
Checking these solutions will give us the unique solution: $$y_0=\frac{6-4\sqrt{21}}{25}$$
$$q.e.d.$$

Can the above method be generalised? Has it been generalised?


A different approach:
You can trade the trigonometric functions for a rational expression with
$$3\sin x+4\cos x=3\frac{2t}{1+t^2}+4\frac{1-t^2}{1+t^2}=2.$$
Thus, the quadratic equation
$$6t^2-6t-2=0,$$ which you can readily solve.
Then
$$\tan x=\frac{2t}{1-t^2}$$ 
and 
$$x=\arctan\frac{-12\pm2\sqrt{21}}5+k\pi.$$
To determine $k$, you must ensure that the angle lies in the quadrant of $(1-t^2,2t)$.

To avoid the computation of $\dfrac{2t}{1-t^2}$, you can also use
$$\tan x=\tan2\frac x2=\frac{2\tan\frac x2}{1-\tan^2\frac x2}=\frac{2t}{1-t^2}$$ or
$$\tan\frac x2=t,$$
$$x=2\arctan t+2k\pi.$$
