Question:

Step 1: Express the right-hand side in terms of a single trigonometric function. We start by rewriting $\sin^2\theta + 2\cos\theta + 1$ using the identity $\sin^2\theta = 1 - \cos^2\theta$: $$f(\theta) = \sin^2\theta + 2\cos\theta + 1 = (1-\cos^2\theta) + 2\cos\theta + 1 = -\cos^2\theta + 2\cos\theta + 2$$ Step 2: Find the range of the expression by substitution. Let $t = \cos\theta$. Since $\theta \in \mathbb{R}$, we have $t \in [-1, 1]$. The expression becomes: $$g(t) = -t^2 + 2t + 2$$ We can rewrite this in vertex form by completing the square: $$g(t) = -(t^2 - 2t) + 2 = -(t^2 - 2t + 1 - 1) + 2 = -(t-1)^2 + 3$$ Step 3: Determine the minimum and maximum values of $g(t)$ on $[-1, 1]$. The function $g(t) = -(t-1)^2 + 3$ is a downward-opening parabola with vertex at $t = 1$. - **Maximum:** At $t = 1$: $g(1) = -(1-1)^2 + 3 = 3$ - **Minimum:** At $t = -1$: $g(-1) = -(-1-1)^2 + 3 = -4 + 3 = -1$ Therefore, the range of $f(\theta)$ is $[-1, 3]$. Step 4: Interpret the condition for the equation to hold for all $\theta$. For the equation $\dfrac{p-1}{2p+3} = \sin^2\theta + 2\cos\theta + 1$ to hold for all $\theta \in \mathbb{R}$, the constant $\dfrac{p-1}{2p+3}$ must lie within the range of the right-hand side. Thus: $$\dfrac{p-1}{2p+3} \in [-1, 3]$$ This gives us the compound inequality: $$-1 \leq \dfrac{p-1}{2p+3} \leq 3$$ Step 5: Solve the left inequality $\dfrac{p-1}{2p+3} \geq -1$. Rearranging: $$\dfrac{p-1}{2p+3} + 1 \geq 0$$ $$\dfrac{p-1 + 2p+3}{2p+3} \geq 0$$ $$\dfrac{3p+2}{2p+3} \geq 0$$ This rational expression is non-negative when both numerator and denominator have the same sign: - **Both positive:** $3p + 2 \geq 0$ and $2p + 3 > 0$ $\Rightarrow$ $p \geq -\dfrac{2}{3}$ and $p > -\dfrac{3}{2}$ $\Rightarrow$ $p \geq -\dfrac{2}{3}$ - **Both negative:** $3p + 2 \leq 0$ and $2p + 3 < 0$ $\Rightarrow$ $p \leq -\dfrac{2}{3}$ and $p < -\dfrac{3}{2}$ $\Rightarrow$ $p < -\dfrac{3}{2}$ Therefore: $p \in \left(-\infty, -\dfrac{3}{2}\right) \cup \left[-\dfrac{2}{3}, \infty\right)$ Step 6: Solve the right inequality $\dfrac{p-1}{2p+3} \leq 3$. Rearranging: $$\dfrac{p-1}{2p+3} - 3 \leq 0$$ $$\dfrac{p-1 - 3(2p+3)}{2p+3} \leq 0$$ $$\dfrac{p - 1 - 6p - 9}{2p+3} \leq 0$$ $$\dfrac{-5p - 10}{2p+3} \leq 0$$ $$\dfrac{-5(p+2)}{2p+3} \leq 0$$ Multiplying by $-1$ (and reversing the inequality): $$\dfrac{5(p+2)}{2p+3} \geq 0$$ This rational expression is non-negative when both numerator and denominator have the same sign: - **Both positive:** $p + 2 \geq 0$ and $2p + 3 > 0$ $\Rightarrow$ $p \geq -2$ and $p > -\dfrac{3}{2}$ $\Rightarrow$ $p > -\dfrac{3}{2}$ - **Both negative:** $p + 2 \leq 0$ and $2p + 3 < 0$ $\Rightarrow$ $p \leq -2$ and $p < -\dfrac{3}{2}$ $\Rightarrow$ $p \leq -2$ Therefore: $p \in (-\infty, -2] \cup \left(-\dfrac{3}{2}, \infty\right)$ Step 7: Find the intersection of both conditions. We need the intersection of: - Case 1: $p \in \left(-\infty, -\dfrac{3}{2}\right) \cup \left[-\dfrac{2}{3}, \infty\right)$ - Case 2: $p \in (-\infty, -2] \cup \left(-\dfrac{3}{2}, \infty\right)$ Computing the intersection: