Check the interval

Question

The real numbers y and k are such that $y = \sec 2x = k ^ 2 - 3k + 1$. Thus, $k$ can take all values of:

Answer: $[ \frac{7}{2}; 4 ]$

Why wouldn't it be $[2; 3]$ ? The intervals of $k$ would be $k\leq 0$ and $k \geq 1$

Answer 1

$\sec2x\in (- \infty, -1]\cup[1, \infty)$

Therefore, $y\geq1$, or $y≤-1$

For $k^2-3k+1\geq1$, $k^2-3k\geq0$ or $k\in(- \infty, 0]\cup[3, \infty)$.

For $k^2-3k+1≤-1$, $k^2-3k+2\leq0$ or $k∈[1,2]$.

$\therefore$ The range of values of k is $k\in(- \infty, 0]\cup [1,2]\cup[3, \infty)$