Question

If k = 4 sin 3X - 5 , then k ∈

Answer 1

Step-by-step explanation:

sin9x+sin3x=0,x∈[0,2π]

⇒3sin3x−4sin

3

3x+sin3x=0

⇒4sin3x−4sin

3

3x=0

⇒4sin3x(1−sin

2

3x)=0

⇒sin3x=0,(1−sin

2

3x)=0

⇒sin3x=0,sin

2

3x=1

⇒3x=nπ,n∈I or 3x=kπ+

2

π

,k∈I

⇒x=

3

nπ

,x=

3

kπ

+

6

π

,k∈I

For n=0,1,2,3,4,5,6 we have

x=0,

3

π

,

3

2π

,π,

3

4π

,

3

5π

,2π

For k=0,1,2,3,4,5 we have

x=

6

π

,

3

π

+

6

π

,

3

2π

+

6

π

,π+

6

π

,

3

4π

+

6

π

,

3

5π

+

6

π

or x=

6

π

,

6

2π

+

6

π

,

6

4π

+

6

π

,

6

6π+π

,

6

8π

+

6

π

,

6

10π

+

6

π

or x=

6

π

,

6

3π

,

6

5π

,

6

7π

,

6

9π

,

6

11π

or x=

6

π

,

2

π

,

6

5π

,

6

7π

,

2

3π

,

6

11π

Thus, the solutions are {0,

3

π

,

3

2π

,π,

3

4π

,

3

5π

,2π,

6

π

,

2

π

,

6

5π

,

6

7π

,

2

3π

,

6

11π

}

or {0,

2

π

,

6

π

,

3

π

,

3

2π

,

2

3π

,π,

3

4π

,

3

5π

,

6

5π

,

6

7π

,

6

11π

,2π}

=13 solutions