Question

The general solution of the system of equations $\left(\text{sin}\right)^{3} \text{x} + \left(\text{sin}\right)^{3} \left(\frac{2 \pi }{3} + \text{x}\right) + \left(\text{sin}\right)^{3} \left(\frac{4 \pi }{3} + \text{x}\right) + \frac{3}{4} \text{cos } 2 \text{x} = 0$ and $cos x\neq 0$ is

• A. $x=\frac{\left(2 \text{k} + 1\right) \pi }{10}\text{, k}\in \text{Z}$
• B. $x=\frac{\left(2 \text{k} + 1\right) \pi }{5}\text{, k}\in \text{Z}$
• C. $x=\frac{\left(4 \text{k} + 1\right) \pi }{10}\text{, k}\in \text{Z}$
• D. $x=\left(\frac{4 \text{k} + 1}{5}\right)\pi \text{, k}\in \text{Z}$

Answer 1

Answer: (C)

$\because sin3x=3sinx-4 \, sin^{3}x\Rightarrow sin^{3}x=\frac{1}{4}\left[3 sinx - sin 3 x \, \right] \, $
The given equation reduces to
$\frac{1}{4} (3 sinx - sin 3 x)+\frac{1}{4} (3 sin (\frac{2 \pi }{3} + x) - \, sin (2 \pi + 3 x )+\frac{1}{4}(3sin(\frac{4 \pi }{3} + x)-sin(4 \pi + 3 x)+\frac{3}{4}cos2x=0 \, $
$3\left\{sin x + sin \left(\frac{2 \pi }{3} + x\right) + sin ⁡ \left(\frac{4 \pi }{3} + x\right)\right\}-\left\{sin ⁡ 3 x + sin ⁡ \left(2 \pi + 3 x\right) + sin ⁡ \left(4 \pi + 3 x\right)\right\}+3cos⁡2x=0$
$3\left\{ \, sinx + 2 sinx cos \, \frac{2 \pi }{3}\right\}-3sin3x+3cos2x=0$
$\Rightarrow \text{sin }3\text{x}=\text{cos }2\text{x }\Rightarrow \text{cos}\left(\frac{\pi }{2} - 3 \text{x}\right)=\text{cos }2\text{x}$ $\Rightarrow 2x=2k\pi \pm\left(\frac{\pi }{2} - 3 x\right)$
$\Rightarrow 5\text{x}=2\text{k}\pi +\frac{\pi }{2}=\left(4 \text{k} + 1\right)\frac{\pi }{2}$ $\forall k\in Z$ or $ \, x=-2k\pi +\frac{\pi }{2}, \, \forall k\in Z$
$\Rightarrow \text{x}=\left(4 \text{k} + 1\right)\frac{\pi }{10}\text{, }\forall k\in Z$ $ \, \left\{cos x \neq 0\right\}$