Question

Solve the following equations.
$sinx+sin2x+sin3x+sin4x=0.$

Answer 1

$sinx+sin2x+sin3x+sin4x=0$

$\Rightarrow (sin4x+sinx)+(sin3x+sin2x)=0$

$\because$ we know,

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

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

$\Rightarrow 2sin\left(\frac{5x}{2}\right)cos\left(\frac{3x}{2}\right)+2sin\left(\frac{5x}{2}\right)cos\left(\frac{x}{2}\right)$

$\Rightarrow 2sin\left(\frac{5x}{2}\right)[cos\left(\frac{3x}{2}\right)+cos\left(\frac{x}{2}\right)]=0$

also we know

$\{cosC+cosD=2cos\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)\}$

$\Rightarrow 2sin\left(\frac{5x}{2}\right)\left[2cos\left(\frac{3/2^x+x/2}{2}\right)cos\left(\frac{3x/2-x/2}{2}\right)\right]=0$

$\Rightarrow 4sin\left(\frac{5x}{2}\right)cosx.cos\left(\frac{x}{2}\right)=0$

Now $sin\left(\frac{5x}{2}\right)=0$ or $cos=0$ or $cos\left(\frac{x}{2}\right)=0$

we know, that General solution for

$sin\theta =0\Rightarrow \{\theta =n\pi ,nϵZ\}.(Z\to$ integer $[0,\pm 1,\pm 2,\pm 3,\pm ...])$

$\therefore forsin\left(\frac{5x}{2}\right)\Rightarrow \{\frac{5x}{2}=n\pi ,nϵZ\}$

$sin(\frac{5x}{2}\Rightarrow \{x=\frac{2}{5}n\pi ,nϵZ\})$

also we know General solution for

$cos\theta =0\Rightarrow \{\theta =(2n+1)\pi /2,nϵZ\}$

For $cosx=0\Rightarrow \{x=(2n+1)\pi /2,nϵZ\}$

for $cos\left(\frac{x}{2}\right)=0\Rightarrow \{\frac{x}{2}=(2n+1),nϵZ\}$

$cos\left(\frac{x}{2}\right)\Rightarrow \{x=(2n+1)\pi ,nϵZ\}$