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

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

.

idk sry

step-by-step explanation:

answer: on graphs, the open and closed circles, or vertical asymptotes drawn as dashed lines us identify discontinuities.

as before, graphs and tables allow us to estimate at best.

when working with formulas, getting zero in the denominator indicates a point of discontinuity.

when working with piecewise-defined functions, check for discontinuities at the transition points where one piece ends and the next begins.

step-by-step explanation: