Question

(Sin4x+2Sin3x+Sin2x)/Sinx=2Sin3xCotx/2​

Answer 1

sorry mate don't know the answer

Answer 2

Answer:

Consider the given equation.

4sinxsin2xsin4x=sin3x

2(2sin2xsinx)sin4x=sin3x

We know that

2sinAsinB=cos(A−B)−cos(A+B)

Therefore,

2[cos(2x−x)−cos(2x+x)]sin4x=sin3x

2[cosx−cos3x]sin4x=sin3x

2sin4xcosx−2sin4xcos3x=sin3x

We know that

2sinAcosB=sin(A+B)+sin(A−B)

Therefore,

sin(4x+x)+sin(4x−x)−[sin(4x+3x)+sin(4x−3x)]=sin3x

sin(5x)+sin(3x)−[sin(7x)+sin(x)]=sin3x

sin(5x)−sin(7x)=sin(x)

We know that

sinC−sinD=2cos(

2

C+D

)sin(

2

C−D

)

Therefore,

2cos(

2

5x+7x

)sin(

2

5x−7x

)=sinx

2cos(

2

12x

)sin(

2

−2x

)=sinx

2cos(6x)sin(−x)=sinx

−2cos(6x)sin(x)=sinx

cos6x=−

2

1

6x=2nπ±

3

2π

x=

3

nπ

±

9

π

Hence, the value is

3

nπ

±

9

π

.