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: $$\begin{align}3\sin x + 4\cos x &= \sqrt{3^2 + 4^2 }\sin\left(x+\arctan\frac{4}{3}\right) \\ 
&= 5\sin(x+\arctan\frac{4}{3})\end{align}$$

In general, if $$a\sin x + b\cos x = c$$
$$\frac{a\sin x}{\sqrt{a^2 + b^2}}+\frac{b \cos x}{\sqrt{a^2 + b^2}} = \frac{c}{\sqrt{a^2 + b^2}}$$
Let $\cos\phi = \dfrac{a}{\sqrt{a^2 + b^2}} \ \ \therefore \sin\phi = \dfrac{b}{\sqrt{a^2 + b^2}}$
Hence $\tan\phi = \dfrac{b}{a}$. Substituting in above eqn,
$$\sin x \cos \phi + \cos x \sin\phi = \frac{c}{\sqrt{a^2 + b^2}}$$
$$\sin (x + \phi)\sqrt{a^2 + b^2} = c$$
$$\implies c = \sqrt{a^2 + b^2}\ \sin\left(x + \arctan\frac{b}{a}\right)$$
A: 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
$$\tan\frac x2=t,$$
$$x=2\arctan t+2k\pi.$$
A: By complex numbers:
By the complex definition of the trigonometric functions, setting $z=e^{ix}$,
$$3\frac{z+z^{-1}}{2i}+4\frac{z+z^{-1}}2=2.$$
Multiplying by $z$ and rearranging,
$$\left(2+i\frac32\right)z^2-2z+\left(2-i\frac32\right)=0.$$
Then solving the quadratic equation
$$z=\frac{(8+6i)\pm\sqrt{21}(3-4i)}{25}.$$
The real and imaginary parts give $\cos x$ and $\sin x$ and $x$ is the imaginary part of the logarithm of $z$.
