$$\begin{align} 2{\sqrt 2}\sin(\phantom{2}x) +3\cos(\phantom{2}y) &= \phantom{-}3.5 \\ 2\sin(2x)+5\cos(2y)&=-0.5 \end{align}$$
I've gotten to the point where I've gotten an equation for $\sin x$ and $\cos x$ using the double angle formulae, but they're horrible roots of quadratics and I don't believe it to be a good method, how else could I attempt this question?
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Consider the system $$\begin{align} a\phantom{2}\sin x +b\phantom{2}\cos y &= p \tag{1}\\ c\sin 2x + d \cos 2y &= q \tag{2} \end{align}$$ With $\sin 2x=2\sin x\cos x$ and $\cos 2y=2\cos^2y-1$, we can write $(2)$ as $$\begin{align} 2c\sin x\cos x= q - d ( 2\cos^2 y - 1 ) \tag{3} \end{align}$$ Squaring both sides and writing $\cos^2 x$ as $1-\sin^2 x$ gives $$4 c^2 \sin^2x(1-\sin^2 x)= \left(\;q-d(2\cos^2 y-1)\;\right)^2 \tag{4}$$
Now, we can use $(1)$ to eliminate $\sin x$ from $(4)$. There's no real benefit in continuing symbolically, so I'll give the result of substituting the specific constants $a=2\sqrt{2}$, $b=3$, $c=2$, $d=5$, $p=7/2$, $q=-1/2$. Conveniently, we have some factorization: $$(2\cos y - 1) \left(3848 \cos^3y- 1100\cos^2 y + 1286\cos y- 2129 \right) = 0 \tag{5}$$ (Note that this would not be so evident if we had eliminated $\cos y$ to get a quartic in $\sin x$.) So, one solution arises from
$$\cos y = \frac12\quad\to\quad y = \pm\frac13\pi \quad\to\quad \begin{cases} 2\sqrt{2}\sin x &= 2 \;(\text{from} (1)) \\ 2 \sin2x &= 2 \;(\text{from} (2)) \end{cases} \quad\to\quad x = \frac14 \pi \tag{6}$$
(with appropriate additions or adjustments for the domain of interest). For the remaining roots, we can use numerical methods (or Mathematica's Solve function) on the cubic factor of $(5)$. Two resulting values of $\cos y$ are non-real; the third is $\cos y = 0.7752\ldots$, and we deduce
$$y = \pm0.6836\ldots \qquad x=2.713\ldots \tag{7}$$
For sure, this is an ugly problem and a numerical method should be used.
What I did was to take $y$ form the first equation $$\cos(y)=\frac{1}{3} \left(\frac{7}{2}-2 \sqrt{2} \sin (x)\right)\implies y=\pm\cos ^{-1}\left(\frac{1}{6} \left(7-4 \sqrt{2} \sin (x)\right)\right)$$ Replace $y$ in the second equation and plot $$f(x)=2\sin(2x)+5\cos(2y)+0.5$$ For the range $0 \leq x \leq 2\pi$, roots look to be close to $x=\frac \pi 4$ and to $x=\pi$. In fact $f\left(\frac{\pi }{4}\right)=0$; so we have one root.
For the second root, we could expand $f(x)$ as a Taylor series built at $x=\pi$ and get $$f(x)=\frac{82}{9}+\left(4+\frac{140 \sqrt{2}}{9}\right) (x-\pi )+\frac{80}{9} (x-\pi )^2+O\left((x-\pi )^3\right)$$ Ignoring the higher order terms and solving the quadratic leads to the approximation $$x = \pi-\frac{5 \sqrt{2}-3}{10} \approx 2.73449$$ while the "exact" solution is $2.71351$.
For sure, we could be better using on more term in the Taylor expansion to get $$f(x)=\frac{82}{9}+\left(4+\frac{140 \sqrt{2}}{9}\right) (x-\pi )+\frac{80}{9} (x-\pi )^2-\frac{2\left(36+35 \sqrt{2}\right)}{27} (x-\pi )^3+O\left((x-\pi )^4\right)$$ and solve this nasty cubic equation (awful expressions for the three real roots) and the solution of interest is $2.70626$ which is better but at the price of a small nightmare.
Using Newton method with $x_0=\pi$ would give the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 3.14159 \\ 1 & 2.79115 \\ 2 & 2.71796 \\ 3 & 2.71353 \\ 4 & 2.71351 \end{array} \right)$$