Question

sinx + sin3x +sin5x=0 give me general solution plzz​

Answer 1

Answer:

• sinx +sin3x+sin5x=0
• added L.H.S
• Sin9x=0
• 1/cosec9x =0
• 1/cosecx=0*9
• 1/cosecx=0
• sinx=0

Answer 2

Step-by-step explanation:

sinx+sin3x+sin5x=0

⇒(sin5x+sinx)+sin3x=0

⇒2sin3xcos2x+sin3x=0

⇒sin3x(2cos2x+1)=0

sin3x=0 or 2cos2x+1=0

sin3x=0 or, cos2x=−  

2

1

​  

 

Now, sin3x=0⟹3x=nπ⟹x=  

3

nπ

​  

,n∈Z

And, cos2x=−  

2

1

​  

 

cos2x=cos  

3

2π

​  

 

2x=2mπ±  

3

2π

​  

,m∈Z

x=mπ±  

3

π

​  

,m∈Z

Hence, the general solution of the given equation is :

x=  

3

nπ

​  

 or, x=mπ±  

3

π

​  

, where m,n∈Z