Question

please solve it ............​

Answer 1

Answer:

★ Graph is Attached

Sin2x-sin4x+sin6x=0

(sin6x+sin2x)-sin4x=0

• We know that sinx +siny=2sin(x+Y/2)cos(x-y/2)

Therefore ,

=2sin(6x+2x/2)cos(6x-2x/2)-sin4x=0

=2sin8x/2cos4x/2-sin4x

sin4x(2cos2x-1)=0

Therefore,

sin4x=0 or 2cos2x-1=0

sin4x=0 or 2cos2x=1

sin4x=0 or cos2x=1/2

• Now

Let sinx=siny

sin4x=sin4y

now sin4x=0

Therfore sin4y=0

sin4y=sin(0)

4y=0

y=0

Now cos2x=1/2

Let cosx=cosy

Cos2x=cos2y

cos2y=1/2 ;cos(2y)=π/3;2y=π/3;

Now,

for sin4x=sin4y the general solution will be

4x=nπ±(-1)ⁿ4y

Putting y=0

4x=nπ

x=nπ/4

For cos2x=cos2y is

2x=2nπ±2y

now 2y=π/3

2x=2nπ±π/3

on solving we get

x=nπ±π/6

Therfore for sin4x=0 ,x=nπ/4

and

for cos2x=1/2,x=nπ±π/6

Answer 2

REFER THE ATTACHMENTS OF THE FULL EXPLANATION