Krdytkyu

$$begin{align}
2{sqrt 2}sin(phantom{2}x) +3cos(phantom{2}y) &= phantom{-}3.5 \
2sin(2x)+5cos(2y)&=-0.5
end{align}$$

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?

Consider the system
$$begin{align}
aphantom{2}sin x +bphantom{2}cos y &= p tag{1}\
csin 2x + d cos 2y &= q tag{2}
end{align}$$
With $sin 2x=2sin xcos x$ and $cos 2y=2cos^2y-1$, we can write $(2)$ as
$$begin{align}
2csin xcos x= q - d ( 2cos^2 y - 1 ) tag{3}
end{align}$$
Squaring both sides and writing $cos^2 x$ as $1-sin^2 x$ gives
$$4 c^2 sin^2x(1-sin^2 x)= left(;q-d(2cos^2 y-1);right)^2 tag{4}$$

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 $a=2sqrt{2}$, $b=3$, $c=2$, $d=5$, $p=7/2$, $q=-1/2$. Conveniently, we have some factorization:
$$(2cos y - 1) left(3848 cos^3y- 1100cos^2 y + 1286cos y- 2129 right) = 0 tag{5}$$
(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

$$cos y = frac12quadtoquad y = pmfrac13pi quadtoquad
begin{cases}
2sqrt{2}sin x &= 2 ;(text{from} (1)) \
2 sin2x &= 2 ;(text{from} (2))
end{cases} quadtoquad x = frac14 pi tag{6}$$

(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 $cos y = 0.7752ldots$, and we deduce

$$y = pm0.6836ldots qquad x=2.713ldots tag{7}$$

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
$$cos(y)=frac{1}{3} left(frac{7}{2}-2 sqrt{2} sin (x)right)implies y=pmcos ^{-1}left(frac{1}{6} left(7-4 sqrt{2} sin (x)right)right)$$ Replace $y$ in the second equation and plot
$$f(x)=2sin(2x)+5cos(2y)+0.5$$ For the range $0 leq x leq 2pi$, roots look to be close to $x=frac pi 4$ and to $x=pi$. In fact $fleft(frac{pi }{4}right)=0$; so we have one root.

For the second root, we could expand $f(x)$ as a Taylor series built at $x=pi$ and get
$$f(x)=frac{82}{9}+left(4+frac{140 sqrt{2}}{9}right) (x-pi )+frac{80}{9} (x-pi
)^2+Oleft((x-pi )^3right)$$ 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
$$f(x)=frac{82}{9}+left(4+frac{140 sqrt{2}}{9}right) (x-pi )+frac{80}{9} (x-pi
)^2-frac{2left(36+35 sqrt{2}right)}{27} (x-pi )^3+Oleft((x-pi )^4right)$$ 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
$$left(
begin{array}{cc}
n & x_n \
0 & 3.14159 \
1 & 2.79115 \
2 & 2.71796 \
3 & 2.71353 \
4 & 2.71351
end{array}
right)$$