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.

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.

answer: okw

step-by-step explanation: