2 sin^2 (x) + 3 sin (x) + 1 = 0
help me find the solution

Step-by-step explanation:

\[2~ sin^2 x+3sin x+1=0\]

\[2sin^2x+2sin x+sin x+1=0\]

2sinx(sin x+1)+1(sin x+1)=0

(sin x+1)(2 sin x+1)=0

either sin x+1=0

sin x=-1=sin 3π/2=sin (2nπ+3π/2)

x=2nπ+3π/2,where n is an integer.

or 2sin x+1=0

sin x=-1/2=-sin π/6=sin (π+π/6),sin (2π-π/6)=sin (2nπ+7π/6),sin (2nπ+11π/6)

x=2nπ+7π/6,2nπ+11π/6,

where n is an integer.

to rewrite a literal equation you may need to divide by a number, or by a variable.

hope i .