# Find the critical numbers of the function. (Use n to denote any arbitrary integer values.) ...

## Question:

Find the critical numbers of the function. (Use n to denote any arbitrary integer values.)

{eq}f(\theta)= 12\cos \theta + 6\sin 2\theta {/eq}

## Critical Points:

Critical points, also known as critical numbers, describe the inputs that lead to the derivative of the function having a value of {eq}0 {/eq} upon the evaluation of the input value in the derivative.

Given: {eq}f(\theta)= 12\cos(\theta)+6\sin(2\theta) {/eq}

To find the critical numbers, the strategy is to set the derivative of the function equal to {eq}0 {/eq} and solve for {eq}\theta {/eq} when this occurs.

{eq}\begin{align*} \frac{d}{d\theta}(f(\theta)) = 0 &\Rightarrow \frac{d}{d\theta}(12\cos(\theta)+6\sin(2\theta)) = 0 \\ &\Rightarrow -12\sin(\theta)+\frac{d}{d\theta}(6\sin(2\theta))\frac{d}{d?}(2?) = 0 \\ &\Rightarrow -12\sin(\theta)+6\cos(2\theta)(2) = 0 \\ &\Rightarrow -12\sin(\theta)+12\cos(2\theta) = 0 \\ &\Rightarrow -12\sin(\theta)+12(1-2\sin^2(\theta)) = 0 \text{ [Trig Identity]} \\ &\Rightarrow -12\sin(\theta)+12-24\sin^2(\theta) = 0 \\ &\Rightarrow -24x^2-12x+12 = 0 \text{ where } x = \sin(\theta) \\ &\Rightarrow -24x^2-12x+12 = 0 \\ &\Rightarrow (-24x+12)(x+1) = 0 \\ &\Rightarrow -24x+12 = 0, x+1 = 0 \\ &\Rightarrow -24x+12-12 = 0-12, x+1-1 = 0-1 \\ &\Rightarrow -24x = -12, x = -1 \\ &\Rightarrow -\frac{1}{24}\cdot -24x = -12\cdot -\frac{1}{24}, x = -1 \\ &\Rightarrow x = \frac{1}{2}, x = -1 \\ &\Rightarrow \sin(\theta) = \frac{1}{2}, \sin(\theta) = -1 \\ &\Rightarrow \sin^{-1}(\sin(\theta)) = \sin^{-1}(\frac{1}{2}), \sin^{-1}(\sin(\theta)) = \sin^{-1}(-1) \\ &\Rightarrow \theta = \frac{\pi}{6}+2\pi n, \frac{5\pi}{6}+2\pi n, \theta = \frac{3\pi}{2}+2\pi n \text{ where } 2\pi n \text{ added to show additional occurrences with the same situation} \\ &\Rightarrow \theta = \frac{\pi}{6}+2\pi n, \frac{5\pi}{6}+2\pi n, \frac{3\pi}{2}+2\pi n \text{ [All ciritical numbers]} \\ \end{align*} {/eq} 