Answer:
The critical value of [tex]f(\theta) = 6\cdot \cos \theta + 3\cdot \sin 2\theta[/tex] are given by [tex]\theta \approx 0.091\pi \pm 2\pi\cdot n[/tex] or [tex]\theta \approx 0.909\pi \pm 2\pi \cdot n[/tex], [tex]\forall \,n \in \mathbb{N}[/tex]
Step-by-step explanation:
The function to be evaluated is [tex]f(\theta) = 6\cdot \cos \theta + 3\cdot \sin 2\theta[/tex], the first derivative of the function must be taken in order to determine the set of critical numbers. Each derivative are found by using the differentiation rule for a sum of functions and rule of chain and subsequently simplified by trigonometric and algebraic means:
First derivative
[tex]f'(\theta) = - 6 \cdot \sin \theta +6\cdot \cos 2\theta[/tex]
[tex]f'(\theta) = -6\cdot \sin \theta + 2\cdot (\cos^{2}\theta-\sin^{2}\theta)[/tex]
[tex]f'(\theta) = -6\cdot \sin \theta + 2\cdot [(1-\sin^{2}\theta-\sin^{2}\theta)][/tex]
[tex]f'(\theta) = -6\cdot \sin \theta + 2\cdot (1-2\cdot \sin^{2}\theta)[/tex]
[tex]f'(\theta) = -6\cdot \sin \theta + 2 - 4\cdot \sin^{2}\theta[/tex]
[tex]f'(\theta) = -4\cdot \sin^{2}\theta - 6\cdot \sin \theta +2[/tex]
The procedure to determine the critical number of the given function are described briefly:
1) First derivative is equalised to zero.
2) The resultant equation is solved.
Then,
[tex]-4\cdot \sin^{2}\theta - 6\cdot \sin \theta +2 = 0[/tex]
Whose roots are:
[tex]\sin \theta_{1} \approx 0.281[/tex] and [tex]\sin \theta_{2} \approx -1.781[/tex]
The sine function is a continuous function with a range between 1 and -1, so, only the first root offers a realistic solution. In addition, such function is positive at first and second quadrants and has a periodicity of [tex]2\pi[/tex] radians, the family of critical values are determined by the unse of inverse trigonometric functions:
[tex]\theta \approx \sin^{-1} 0.281[/tex]
There are two subsets of solutions:
[tex]\theta \approx 0.091\pi \pm 2\pi\cdot n[/tex] or [tex]\theta \approx 0.909\pi \pm 2\pi \cdot n[/tex], [tex]\forall \,n \in \mathbb{N}[/tex]
