Question

solve sin 2A + sin 4A + sin 6 A =0

Answer 1

hope this will help you:-)

Answer 2

Solved, sin 2A + sin 4A + sin 6 A = 0.

Given

To solve sin 2A + sin 4A + sin 6A = 0.

Formula's used in the problem :

     Sin C + Sin D = 2 Sin$(\frac{C+D}{2})$ Cos$(\frac{C-D}{2})$

     Cos 2A = 2$Cos^2A - 1$

Steps :              

            Sin 2A + Sin 4A + Sin 6A = 0    

            (Sin 6A + Sin 2A) +Sin 4A =0    

Using the formula, "Sin C + Sin D = 2 Sin$(\frac{C+D}{2})$ Cos$(\frac{C-D}{2})$ "        

             $\begin{equation}2 \sin \left(\frac{6 A+2 A}{2}\right) \cos \left(\frac{6 A-2 A}{2}\right)+\sin 4 A=0\end$      

             2 Sin 4A. Cos 2A + Sin 4A = 0

              Sin 4A ( 2Cos 2A + 1 ) = 0

              Sin 4A = 0   ;   2 Cos 2A + 1 = 0

Sin 4A = 0

Hence, Sine is “O” which is at the integral multiple of π.

       4A = nπ

         A = nπ/4

Where, n = integer.

2 Cos 2A + 1 = 0                      

            2 Cos 2A = -1

            Cos 2A = (-1)/2

Using formula, “ Cos 2A = 2 $Cos^2A - 1$ "

            2 $Cos^2$ A -1 = -1/2

            2 cos² A = 1 - 1/2

            2 cos² A = 1/2

               cos² A = 1/4

               cos² A = (1/2 )²

               cos² A = cos² π/3

General solution for,

               cos² x = cos² y

               x = nπ ± y                 A = nπ ± π/3      

               A = π/3 ( 3n±1 )

               n = integer.

Therefore, the problem sin 2A + sin 4A + sin 6 A = 0 has solved.

To learn more...

1. brainly.in/question/6820230

2. brainly.in/question/752610