Question

solve the eqn :
4sin x sin 2x sin 4x = sin 3x​

Answer 1

$\huge\green{\mathfrak{\underline{\underline{ANSWER:-}}}}$

$4 \sin(x) \sin(2x) \sin(4x) = \sin(3x ) \\ 2(2 \sin(2x) \sin(x) ) \sin(4x) = \sin(3x)$

As we know that:-

$2 \sin(a) \ \sin(b) = \cos(a - b) - \cos(a + b)$

$= > 2( \cos(2x - x) - \cos(2x + x) ) \sin(4x) = \sin(3x) \\ 2( \cos(x) - \cos(3x) ) \sin(4x) = \sin(3x) \\ 2 \sin(4x ) \cos(x) - 2 \sin(4x) \cos(3x) = \sin(3x)$

Also,

$2 \sin(a) \cos(b) = \sin(a + b) + \sin(a - b)$

$= > \sin(4x + x) + \sin(4x - x) - \sin(4x + 3x) - \sin(4x - 3x) = \sin(3x) \\ \sin(5x) + \sin(3x) - \sin(7x) - \sin(x) = \sin(3x) \\ = > \sin(5x) - \sin(7x) = \sin(x)$

Also,

$\sin(c) - \sin(d) = 2 \cos( \frac{c + d}{2} ) \sin( \frac{c - d}{2} )$

$= > 2 \cos( \frac{5x + 7x}{2} ) \sin( \frac{5x - 7x}{2} ) = \sin(x) \\2 \cos( 6 x) \sin( - x) = \sin(x) \\ - 2 \cos(6x) \sin(x) = \sin(x) \\ = > \cos(6x) = - \frac{1}{2}$

$= > 6x = 2n\pi + - \frac{2\pi}{3} \\ x = \frac{n\pi}{3} + - \frac{\pi}{9}$

HOPE THIS HELPS YOU!!!!!!!

PLEASE MARK IT AS BRAINLIST!!!

#STAY HOME #STAY SAFE

Answer 2

\huge\green{\mathfrak{\underline{\underline{ANSWER:-}}}}

ANSWER:−

\begin{gathered}4 \sin(x) \sin(2x) \sin(4x) = \sin(3x ) \\ 2(2 \sin(2x) \sin(x) ) \sin(4x) = \sin(3x) \\\end{gathered}

4sin(x)sin(2x)sin(4x)=sin(3x)

2(2sin(2x)sin(x))sin(4x)=sin(3x)

As we know that:-

2 \sin(a) \ \sin(b) = \cos(a - b) - \cos(a + b)2sin(a) sin(b)=cos(a−b)−cos(a+b)

\begin{gathered}= > 2( \cos(2x - x) - \cos(2x + x) ) \sin(4x) = \sin(3x) \\ 2( \cos(x) - \cos(3x) ) \sin(4x) = \sin(3x) \\ 2 \sin(4x ) \cos(x) - 2 \sin(4x) \cos(3x) = \sin(3x)\end{gathered}

=>2(cos(2x−x)−cos(2x+x))sin(4x)=sin(3x)

2(cos(x)−cos(3x))sin(4x)=sin(3x)

2sin(4x)cos(x)−2sin(4x)cos(3x)=sin(3x)

Also,

\begin{gathered}2 \sin(a) \cos(b) = \sin(a + b) + \sin(a - b) \\\end{gathered}

2sin(a)cos(b)=sin(a+b)+sin(a−b)

\begin{gathered}= > \sin(4x + x) + \sin(4x - x) - \sin(4x + 3x) - \sin(4x - 3x) = \sin(3x) \\ \sin(5x) + \sin(3x) - \sin(7x) - \sin(x) = \sin(3x) \\ = > \sin(5x) - \sin(7x) = \sin(x) \\\end{gathered}

=>sin(4x+x)+sin(4x−x)−sin(4x+3x)−sin(4x−3x)=sin(3x)

sin(5x)+sin(3x)−sin(7x)−sin(x)=sin(3x)

=>sin(5x)−sin(7x)=sin(x)

Also,

\sin(c) - \sin(d) = 2 \cos( \frac{c + d}{2} ) \sin( \frac{c - d}{2} )sin(c)−sin(d)=2cos(

2

c+d

)sin(

2

c−d

)

\begin{gathered}= > 2 \cos( \frac{5x + 7x}{2} ) \sin( \frac{5x - 7x}{2} ) = \sin(x) \\2 \cos( 6 x) \sin( - x) = \sin(x) \\ - 2 \cos(6x) \sin(x) = \sin(x) \\ = > \cos(6x) = - \frac{1}{2} \\\end{gathered}

=>2cos(

2

5x+7x

)sin(

2

5x−7x

)=sin(x)

2cos(6x)sin(−x)=sin(x)

−2cos(6x)sin(x)=sin(x)

=>cos(6x)=−

2

1

\begin{gathered}= > 6x = 2n\pi + - \frac{2\pi}{3} \\ x = \frac{n\pi}{3} + - \frac{\pi}{9}\end{gathered}

=>6x=2nπ+−

3

2π

x=

3

nπ

+−

9

π

HOPE THIS HELPS YOU!!!!!!!

PLEASE MARK IT AS BRAINLIST!!!

#STAY HOME #STAY SAFE

Thanks!