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 $cosx\ne 0$ 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)

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