Question

$\frac{10}{F}=3Sin\theta +4Cos\theta$

Answer 1

$F \in (-\infty, -2] \cup [2, \infty)$

Answer 2

The given equation is: $\frac{10}{F} = 3\sin\theta + 4\cos\theta$

We know that for an expression of the form $a\sin\theta + b\cos\theta$, its range is given by $[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$.
In this case, $a=3$ and $b=4$.
So, $\sqrt{a^2+b^2} = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5$.

Therefore, the range of the expression $3\sin\theta + 4\cos\theta$ is $[-5, 5]$.
This implies: $-5 \le \frac{10}{F} \le 5$ This inequality can be split into two separate inequalities:

1. $\frac{10}{F} \le 5$
2. $\frac{10}{F} \ge -5$

Let's solve the first inequality: $\frac{10}{F} \le 5$ $\frac{10}{F} - 5 \le 0$ $\frac{10 - 5F}{F} \le 0$ $\frac{5(2 - F)}{F} \le 0$ $\frac{2 - F}{F} \le 0$ To solve this, we find the critical points by setting the numerator and denominator to zero, which are $F=2$ and $F=0$. We analyze the sign of the expression in different intervals:

• For $F < 0$: Let $F=-1$. $\frac{2 - (-1)}{-1} = \frac{3}{-1} = -3 \le 0$. This interval is part of the solution.
• For $0 < F < 2$: Let $F=1$. $\frac{2 - 1}{1} = \frac{1}{1} = 1 > 0$. This interval is not part of the solution.
• For $F > 2$: Let $F=3$. $\frac{2 - 3}{3} = \frac{-1}{3} \le 0$. This interval is part of the solution. Also, when $F=2$, the expression is $\frac{0}{2}=0$, which satisfies $\le 0$. So $F=2$ is included.
  Thus, the solution for the first inequality is $F \in (-\infty, 0) \cup [2, \infty)$.

Now, let's solve the second inequality: $\frac{10}{F} \ge -5$ $\frac{10}{F} + 5 \ge 0$ $\frac{10 + 5F}{F} \ge 0$ $\frac{5(2 + F)}{F} \ge 0$ $\frac{2 + F}{F} \ge 0$ The critical points are $F=-2$ and $F=0$. We analyze the sign of the expression in different intervals:

• For $F < -2$: Let $F=-3$. $\frac{2 + (-3)}{-3} = \frac{-1}{-3} = \frac{1}{3} \ge 0$. This interval is part of the solution.
• For $-2 < F < 0$: Let $F=-1$. $\frac{2 + (-1)}{-1} = \frac{1}{-1} = -1 < 0$. This interval is not part of the solution.
• For $F > 0$: Let $F=1$. $\frac{2 + 1}{1} = \frac{3}{1} = 3 \ge 0$. This interval is part of the solution. Also, when $F=-2$, the expression is $\frac{0}{-2}=0$, which satisfies $\ge 0$. So $F=-2$ is included.
  Thus, the solution for the second inequality is $F \in (-\infty, -2] \cup (0, \infty)$.

To find the values of $F$ for which the original equation holds, we need to find the intersection of the solutions from both inequalities: $S_1 = (-\infty, 0) \cup [2, \infty)$ $S_2 = (-\infty, -2] \cup (0, \infty)$

The intersection $S_1 \cap S_2$ is:

• For $F \le -2$: Both $S_1$ and $S_2$ are satisfied. So $(-\infty, -2]$ is part of the solution.
• For $-2 < F < 0$: $S_1$ is satisfied, but $S_2$ is not.
• For $0 < F < 2$: $S_1$ is not satisfied, but $S_2$ is.
• For $F \ge 2$: Both $S_1$ and $S_2$ are satisfied. So $[2, \infty)$ is part of the solution.

Combining these parts, the possible values for $F$ are $F \in (-\infty, -2] \cup [2, \infty)$.
This can also be expressed as $|F| \ge 2$.