I am trying to figure out if there is a mistake in this question or not. The question is as follows
In the markscheme page 24 it is shown that $x= \pm 1$. However, I belive that $x=1$ only. The way I came to this conclusion is that in part (ii) I showed that $y^2+1\geq4(y-1)^2$ by writing $$y\cos \theta-\sin \theta =\frac{y}{|y|}\sqrt(1+y^2)\cos(\theta + \phi)\leq\sqrt{1+y^2}$$ where $\tan (\phi)=\frac{1}{y}$ and then squaring to get the desired inequality. So if we have the scenario in (iii) it is implied that equality occurs and hence solving for $x$ in the original equation for $y$ gives $$x^2(1-y)+x(\cos \theta - y\cos \theta) + 1-y=0$$ $$\implies x=1 \quad \text{using the above for the trig coef of $x$}$$
The markscheme seems to have said that since the given information in $(iii)$ implies equality we have $$(y\cos \theta-\sin \theta)^2 =1+y^2 \implies y\cos \theta-\sin \theta =\pm\sqrt{1+y^2}$$ however, as shown previously, this can only occur when $y<0$ and in this part of the question we are given $y=\frac{4+\sqrt{7}}{3}$ and hence $y>0$ implying the positive root only, and hence, one solution. I went on to plot some graphs on desmos and it seems like there are no such graphs that $y=\frac{4+\sqrt{7}}{3}$ when $x=1$ and $x=-1$. Here is one
I tried plotting a 3d diagram on wolfram to see if I am correct, however, it is still loading.
I am probably mistaken somewhere, but I can't seem to figure out where. If anybody can assist me I will be very thankful.