Question

For what $x$is the derivative of the function $f(x)$equal to zero?
$f(x) = \operatorname{Sin} 3x - \sqrt 3 \cos 3x + 3(\cos x - \sqrt 3 \sin x)$

Answer 1

We will first simplify the given function and use the formulas of Trigonometric Functions.

$\sin (A \pm B) = \sin A\cos B \pm \cos A\sin B$
$\cos (A \pm B) = \cos A\cos B \mp \sin A\sin B$
After that we will find the derivative of the simplified function with respect to $x$. The derivative with respect to $x$is represented as $\dfrac{d}{{dx}}$. We know,
$\dfrac{d}{{dx}}(\sin ax) = a\cos ax$
$\dfrac{d}{{dx}}(\cos ax) = - a\sin ax$
$\dfrac{d}{{dx}}ax = a$
After finding the derivative, we need to equate it to zero and then solve for $x$in that equation.

Complete step-by-step solution:
We are given, $f(x) = \operatorname{Sin} 3x - \sqrt 3 \cos 3x + 3(\cos x - \sqrt 3 \sin x)$
Multiplying and Dividing the right hand side by $2$, we get
$f(x) = \dfrac{2}{2}[\operatorname{Sin} 3x - \sqrt 3 \cos 3x + 3(\cos x - \sqrt 3 \sin x)]$
$f(x) = 2[\dfrac{1}{2}\operatorname{Sin} 3x - \dfrac{{\sqrt 3 }}{2}\cos 3x + \dfrac{3}{2}(\cos x - \sqrt 3 \sin x)]$ (Shifting $\dfrac{1}{2}$inside the bracket)
$f(x) = 2[\dfrac{1}{2}\operatorname{Sin} 3x - \dfrac{{\sqrt 3 }}{2}\cos 3x + 3(\dfrac{1}{2}\cos x - \dfrac{{\sqrt 3 }}{2}\sin x)]$ (Opening Brackets) -----(1)
We know, $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$ and $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$
Substituting it in the above equation, we get
$f(x) = 2[\cos \dfrac{\pi }{3}\operatorname{Sin} 3x - \sin \dfrac{\pi }{3}\cos 3x + 3(\cos \dfrac{\pi }{3}\cos x - \sin \dfrac{\pi }{3}\sin x)]$
Comparing with $\sin (A \pm B) = \sin A\cos B \pm \cos A\sin B$and $\cos (A \pm B) = \cos A\cos B \mp \sin A\sin B$with the above equation, we get
$f(x) = 2[\sin (3x - \dfrac{\pi }{3}) + 3\cos (x + \dfrac{\pi }{3})]$
Now, we need the whole equation either in terms of $\sin x$or $\cos x$
So, we will substitute $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$and $\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}$for the last two terms in (1)
Therefore, we get
$f(x) = 2[\cos \dfrac{\pi }{3}\operatorname{Sin} 3x - \sin \dfrac{\pi }{3}\cos 3x + 3(\sin \dfrac{\pi }{6}\cos x - \cos \dfrac{\pi }{6}\sin x)]$
Now, using the trigonometric formulas, we get
$f(x) = 2[\sin (3x - \dfrac{\pi }{3}) + 3\sin (\dfrac{\pi }{6} - x)]$
Now, finding the derivative of $f(x) = 2[\sin (3x - \dfrac{\pi }{3}) + 3\sin (\dfrac{\pi }{6} - x)]$ with respect to $x$.
$\Rightarrow \dfrac{d}{{dx}}[f(x)] = \dfrac{d}{{dx}}[2[\sin (3x - \dfrac{\pi }{3}) + 3\sin (\dfrac{\pi }{6} - x)]]$
$\Rightarrow \dfrac{d}{{dx}}[f(x)] = 2[\dfrac{d}{{dx}}\sin (3x - \dfrac{\pi }{3}) + 3\dfrac{d}{{dx}}\sin (\dfrac{\pi }{6} - x)]$
$\Rightarrow \dfrac{d}{{dx}}[f(x)] = 2[\cos (3x - \dfrac{\pi }{3}) \times \dfrac{d}{{dx}}(3x) + 3(\cos (\dfrac{\pi }{6} - x) \times \dfrac{d}{{dx}}( - x))]$ (Chain Rule)
$\Rightarrow \dfrac{d}{{dx}}[f(x)] = 2[\cos (3x - \dfrac{\pi }{3}) \times 3 + 3(\cos (\dfrac{\pi }{6} - x) \times ( - 1))]$
$\Rightarrow \dfrac{d}{{dx}}[f(x)] = 2 \times 3[\cos (3x - \dfrac{\pi }{3}) - \cos (\dfrac{\pi }{6} - x)]$ (Taking $3$common)
$\Rightarrow \dfrac{d}{{dx}}[f(x)] = 6 \times [\cos (3x - \dfrac{\pi }{3}) - \cos (\dfrac{\pi }{6} - x)]$
Now, equating this derivative of $f(x)$to zero.
$\Rightarrow \dfrac{d}{{dx}}[f(x)] = 6 \times [\cos (3x - \dfrac{\pi }{3}) - \cos (\dfrac{\pi }{6} - x)] = 0$
$\Rightarrow 6 \times [\cos (3x - \dfrac{\pi }{3}) - \cos (\dfrac{\pi }{6} - x)] = 0$
$[ \cos (3x - \dfrac{\pi }{3}) - \cos (\dfrac{\pi }{6} - x)] = 0$ (Shifting the terms)
$\Rightarrow \cos (3x - \dfrac{\pi }{3}) = \cos (\dfrac{\pi }{6} - x)$ (Shifting) ---(2)
We know, $\cos A = \cos B \Rightarrow A = 2n\pi \pm B,n \in \mathbb{Z}$
$\therefore$Using this in (2), we get
$3x - \dfrac{\pi }{3} = 2n\pi \pm (\dfrac{\pi }{6} - x),n \in \mathbb{Z}$
$\therefore$Considering two cases, we get
CASE 1 - $3x - \dfrac{\pi }{3} = 2{n_1}\pi + (\dfrac{\pi }{6} - x),{n_1} \in \mathbb{Z}$
CASE 2 - $3x - \dfrac{\pi }{3} = 2{n_2}\pi - (\dfrac{\pi }{6} - x),{n_2} \in \mathbb{Z}$
Solving for $x$in both the cases individually.
From CASE 1, we get
$3x - \dfrac{\pi }{3} = 2{n_1}\pi + (\dfrac{\pi }{6} - x)$ ${n_1} \in \mathbb{Z}$
$\Rightarrow 3x - \dfrac{\pi }{3} = 2{n_1}\pi + \dfrac{\pi }{6} - x$ ${n_1} \in \mathbb{Z}$
$\Rightarrow 3x + x = 2{n_1}\pi + \dfrac{\pi }{6} + \dfrac{\pi }{3}$ ${n_1} \in \mathbb{Z}$ (Shifting the terms)
$\Rightarrow 4x = 2{n_1}\pi + \dfrac{{\pi + 2\pi }}{6}$ ${n_1} \in \mathbb{Z}$ (Taking LCM)
$\Rightarrow x = \dfrac{1}{4}[2{n_1}\pi + \dfrac{{3\pi }}{6}]$ ${n_1} \in \mathbb{Z}$ (Dividing by $4$)
$\Rightarrow x = \dfrac{1}{4}[2{n_1}\pi + \dfrac{\pi }{2}]$ ${n_1} \in \mathbb{Z}$ (Simplifying)
From CASE 2, we get
$3x - \dfrac{\pi }{3} = 2{n_2}\pi - (\dfrac{\pi }{6} - x)$ ${n_2} \in \mathbb{Z}$
$\Rightarrow 3x - \dfrac{\pi }{3} = 2{n_2}\pi - \dfrac{\pi }{6} + x$ ${n_2} \in \mathbb{Z}$ (Opening Brackets)
$\Rightarrow 3x - x = 2{n_2}\pi - \dfrac{\pi }{6} + \dfrac{\pi }{3}$ ${n_2} \in \mathbb{Z}$ (Shifting variables on one side)
$\Rightarrow 2x = 2{n_2}\pi + \dfrac{{ - \pi + 2\pi }}{6}$ ${n_2} \in \mathbb{Z}$
$\Rightarrow 2x = 2{n_2}\pi + \dfrac{\pi }{6}$ ${n_2} \in \mathbb{Z}$
$\Rightarrow x = \dfrac{1}{2}[2{n_2}\pi + \dfrac{\pi }{6}]$ ${n_2} \in \mathbb{Z}$ (Dividing by $2$)
Hence, from the above two cases, we got
$x = \dfrac{1}{4}[2{n_1}\pi + \dfrac{\pi }{2}]$ and $x = \dfrac{1}{2}[2{n_2}\pi + \dfrac{\pi }{6}]$ , where ${n_1},{n_2} \in \mathbb{Z}$
Therefore, for $x = \dfrac{1}{4}[2{n_1}\pi + \dfrac{\pi }{2}]$ and $x = \dfrac{1}{2}[2{n_2}\pi + \dfrac{\pi }{6}]$, ${n_1},{n_2} \in \mathbb{Z}$, the derivative for the function $f(x)$is equal to zero, where $f(x) = \operatorname{Sin} 3x - \sqrt 3 \cos 3x + 3(\cos x - \sqrt 3 \sin x)$.

Note: We could have solved the problem without first simplifying $f(x)$but that could be very tough and we might get stuck in between the problem. Also, we need to be very thorough with all the formulas for trigonometry otherwise we won’t be able to solve the problem in this way. And, we usually forget to apply chain rule while differentiating the terms, so it should be kept in mind. Usually, we don’t write ${n_1},{n_2} \in \mathbb{Z}$ but it is important to write.