Question

Solve the following equations.
$sin\frac{2x+1}{x}+sin\frac{2x+1}{3x}-3co{s}^2\frac{2x+1}{3x}=0.$

Answer 1

$sin\frac{2x+1}{x}+sin\frac{2x+1}{3x}-3co{s}^2\frac{2x+1}{3x}=0.$

using $sinC+sinD=2sin\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)$

$\Rightarrow 2sin\frac{1}{2}(\frac{2x+1}{x}+\frac{2x+1}{3x})cos\frac{1}{2}(\frac{2x+1}{x}-\frac{2x+1}{3x})-3co{s}^2\left(\frac{2x+1}{3x}\right)=0$

$\Rightarrow 2sin\left(\frac{6x+3+2x+1}{3x}\right)cos\frac{1}{2}\left(\frac{6x+3-2x-1}{3}\right)-3co{s}^2\left(\frac{2x+1}{3x}\right)=0$

$\Rightarrow 2in\frac{4x+2}{3x}cos\frac{2x+1}{3x}-3co{s}^2\frac{2x+1}{3x}=0$

$\Rightarrow cos\left(\frac{2x+1}{3x}\right)[2sin2x\left(\frac{2x+1}{3x}\right)-3cos\left(\frac{2x+1}{3x}\right)]=0$

$\Rightarrow cos\left(\frac{2x+1}{3x}\right)[4sin\left(\frac{2x+1}{3x}\right)cos\left(\frac{2x+1}{3x}\right)-3cos\left(\frac{2x+1}{3x}\right)]=0$

$\Rightarrow co{s}^2\left(\frac{2x+1}{3x}\right)\left[4sin(\frac{2x+1}{3x}-3)\right]=0$

$\Rightarrow cos\frac{2x+1}{3x}=0$ or $4sin\frac{2x+1}{3x}-3=0$

$\Rightarrow cos\left(\frac{2x+1}{3x}\right)=cos\left(\pi /2\right)$ or $sin\frac{2x+1}{3x}=\frac{3}{4}$

$\Rightarrow \frac{2x+1}{3x}=\frac{\pi }{2}$ or $\frac{2x+1}{3x}=si{n}^{-1}\left(3/4\right)$

$\Rightarrow \frac{2}{3}+\frac{1}{3x}=\frac{\pi }{2}$ or $\frac{2x+1}{3x}=si{n}^{-1}\left(3/4\right)$

$\Rightarrow \frac{1}{3x}=\frac{\pi }{2}-\frac{2}{3}$ or $\frac{2}{3}+\frac{1}{3x}=si{n}^{-1}\left(3/4\right)$

$\Rightarrow \frac{1}{3x}=\frac{3\pi -4}{6}$ or $\frac{1}{3x}=si{n}^{-1}\left(3/4\right)-2/3$

$x=\frac{2}{3x-4}$

$x=\frac{1}{3(si{n}^{-1}3/4-2/3)}$

$x=\frac{1}{3si{n}^{-1}3/4-2}$