Question

At $x = \dfrac{{5\pi }}{6}$ , the value of $2\sin 3x + 3\cos 3x$ is
1. $0$
2. $1$
3. $- 1$
4. None of these

Answer 1

We know that from trigonometric identity,
$\sin \left( {2\pi + \theta } \right) = \sin \theta$
And
$\cos \left( {2\pi + \theta } \right) = \cos \theta$
Given $x = \dfrac{{5\pi }}{6}$
Here we are asked to find the value of $2\sin 3x + 3\cos 3x$
Substitute the value of $x$ in $2\sin 3x + 3\cos 3x$
$\Rightarrow 2\sin 3x + 3\cos 3x = 2\sin \left( {3 \times \dfrac{{5\pi }}{6}} \right) + 3\cos \left( {3 \times \dfrac{{5\pi }}{6}} \right)$
$= 2\sin \left( {\dfrac{{5\pi }}{2}} \right) + 3\cos \left( {\dfrac{{5\pi }}{2}} \right)$
Write the above equation in terms of $\sin \left( {2\pi + \theta } \right)$ and $\cos \left( {2\pi + \theta } \right)$ .
Then simplify the equation to get the value of the given equation.

Complete step-by-step solution:
Given $x = \dfrac{{5\pi }}{6}$
$\therefore 2\sin 3x + 3\cos 3x = 2\sin \left( {3 \times \dfrac{{5\pi }}{6}} \right) + 3\cos \left( {3 \times \dfrac{{5\pi }}{6}} \right)$
$= 2\sin \left( {\dfrac{{5\pi }}{2}} \right) + 3\cos \left( {\dfrac{{5\pi }}{2}} \right)$
Writing the above equation in terms of $\sin \left( {2\pi + \theta } \right)$ and $\cos \left( {2\pi + \theta } \right)$ , we get
$2\sin 3x + 3\cos 3x = 2\sin \left( {2\pi + \dfrac{\pi }{2}} \right) + 3\cos \left( {2\pi + \dfrac{\pi }{2}} \right)$
We know that $\sin \left( {2\pi + \theta } \right) = \sin \theta$ and $\cos \left( {2\pi + \theta } \right) = \cos \theta$
$\therefore 2\sin 3x + 3\cos 3x = 2\sin \left( {\dfrac{\pi }{2}} \right) + 3\cos \left( {\dfrac{\pi }{2}} \right)$
We know that $\sin \left( {\dfrac{\pi }{2}} \right) = 1$ and $\cos \left( {\dfrac{\pi }{2}} \right) = 0$
$\therefore 2\sin 3x + 3\cos 3x = 2 \times 1 + 3 \times 0 = 2$
Hence the value of $2\sin 3x + 3\cos 3x$ is $2$

Note: The formula $\sin \left( {2\pi + \theta } \right) = \sin \theta$ and $\cos \left( {2\pi + \theta } \right) = \cos \theta$ is derived using trigonometric addition formula
We know that $\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b$
$\therefore \sin \left( {2\pi + \theta } \right) = \sin 2\pi \cos \theta + \cos 2\pi \sin \theta$
The value of $\sin 2\pi = 0$ and $\cos 2\pi = 1$
$\Rightarrow \sin \left( {2\pi + \theta } \right) = 0 \times \cos \theta + 1 \times \sin \theta$
$\therefore \sin \left( {2\pi + \theta } \right) = \sin \theta$
Similarly, $\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b$
$\therefore \cos \left( {2\pi + \theta } \right) = \cos 2\pi \cos \theta - \sin 2\pi \sin \theta$
$\Rightarrow \cos \left( {2\pi + \theta } \right) = \cos \theta$