Question

Let $S=\{\theta \in [0,2\pi]: 8^{2\cos^{2}\theta}+8^{2\sin^{2}\theta}=16\}$

• A. 0
• B. -2
• C. -4
• D. 12

Answer 1

Answer: (A)

0

Answer 2

The given equation is $8^{2\cos^{2}\theta}+8^{2\sin^{2}\theta}=16$. Let $x = 2\cos^2\theta$. We know that $\sin^2\theta + \cos^2\theta = 1$, so $2\sin^2\theta = 2(1-\cos^2\theta) = 2 - 2\cos^2\theta = 2-x$. Substitute these into the equation: $8^x + 8^{2-x} = 16$. Let $y = 8^x$. The equation becomes: $y + \frac{8^2}{y} = 16$, $y + \frac{64}{y} = 16$. Multiply by $y$ (assuming $y \neq 0$, which is true since $8^x$ is always positive): $y^2 + 64 = 16y$, $y^2 - 16y + 64 = 0$. This is a perfect square trinomial: $(y-8)^2 = 0$. So, $y=8$. Now substitute back $y = 8^x$: $8^x = 8$. This implies $x=1$. Substitute back $x = 2\cos^2\theta$: $2\cos^2\theta = 1$, $\cos^2\theta = \frac{1}{2}$, $\cos\theta = \pm \frac{1}{\sqrt{2}}$. We need to find $\theta \in [0, 2\pi]$ for which $\cos\theta = \pm \frac{1}{\sqrt{2}}$. The values of $\theta$ in the interval $[0, 2\pi]$ are: 1. When $\cos\theta = \frac{1}{\sqrt{2}}$: $\theta = \frac{\pi}{4}$ (1st quadrant) and $\theta = \frac{7\pi}{4}$ (4th quadrant). 2. When $\cos\theta = -\frac{1}{\sqrt{2}}$: $\theta = \frac{3\pi}{4}$ (2nd quadrant) and $\theta = \frac{5\pi}{4}$ (3rd quadrant). So, the set $S = \left\{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\right\}$. The question is incomplete, stating "then" and providing integer options. A common type of question in such a scenario is to ask for the sum of values of $\cos(2\theta)$ for $\theta \in S$. Let's calculate $\cos(2\theta)$ for each $\theta \in S$. We know $\cos(2\theta) = 2\cos^2\theta - 1$. Since $2\cos^2\theta = 1$ for all $\theta \in S$, we have: $\cos(2\theta) = 1 - 1 = 0$. Therefore, for every $\theta \in S$, $\cos(2\theta) = 0$. The sum of $\cos(2\theta)$ for all $\theta \in S$ would be: $\sum_{\theta \in S} \cos(2\theta) = 0 + 0 + 0 + 0 = 0$. This matches option (1).