To answer this question we will use the following trigonometric identity:
[tex]\sin^2\theta+\cos^2\theta=1.[/tex]
Solving the first equation for cosθ we get:
[tex]\begin{gathered} x-2=4\cos\theta+2-2, \\ x-2=4\cos\theta, \\ \frac{x-2}{4}=\frac{4\cos\theta}{4}, \\ \frac{x-2}{4}=\cos\theta. \end{gathered}[/tex]
Solving the second equation for sinθ we get:
[tex]\begin{gathered} y+5=2\sin\theta-5+5, \\ y+5=2\sin\theta, \\ \frac{y+5}{2}=\frac{2\sin\theta}{2} \\ \frac{y+5}{2}=\sin\theta. \end{gathered}[/tex]
Substituting
[tex]\cos\theta=\frac{x-2}{4}\text{ and }\sin\theta=\frac{y+5}{2}[/tex]
in the trigonometric identity we get:
[tex](\frac{x-2}{4})^2+(\frac{y+5}{2})^2=1.[/tex]
Simplifying the above result we get:
[tex]\frac{(x-2)^2}{16}+\frac{(y+5)^2}{4}=1[/tex]
Finally, recall that:
[tex]-1\leq\cos\theta\le1.[/tex]
Therefore:
[tex]\begin{gathered} -4+2\leq4\cos\theta+2\leq4+2, \\ -2\leq4\cos\theta+2\leq6. \end{gathered}[/tex]
Therefore x is on the interval:
[tex][-2,6].[/tex]
Answer:
[tex]\frac{(x-2)^2}{16}+\frac{(y+5)^2}{4}=1[/tex]
where x is on the interval
[tex][-2,6].[/tex]
