Question

answer...............​

Answer 1

||✪✪ QUESTION ✪✪||

Solve The Equation sinA + sin3A + sin5A = 0

|| ✰✰ ANSWER ✰✰ ||

→ sinA + sin3A + sin5A = 0

→ (sinA + sin5A) + sin3A = 0

Using sinC + sinD = 2 * sin(C+D)/2 * cos(C-D)/2 ,

→ [ 2 * sin(A+5A)/2 * cos(A - 5A)/2 ] + sin3A = 0

→ [ 2 * sin(6A)/2 * cos(-4A)/2 ] + sin3A = 0

→ 2 * sin3A * cos(-2A) + sin3A = 0

Now, using cos(-A) = cosA ,

→ 2 * sin3A * cos2A + sin3A = 0

Taking sin3A common now,

→ sin3A( 2cos2A + 1) = 0

Now, putting both Equal to Zero now, we get,

→ sin3A = 0

→ 3A = nπ

→ A = (nπ)/3 (Ans.)

Or,

→ (2cos2A + 1) = 0

→ 2cos2A = (-1)

→ cos2A = (-1/2)

→ 2A = 2nπ ± (-π)/3

→ A = [ 2nπ ± (-π)/3 ] /2

→ A = nπ ± (-π)/6

→ A = nπ ∓ (π/6) (Ans.)

Answer 2

Step-by-step explanation:

$\huge\underline\purple{\sf Answer :-}$ Please check the attached picture