Answer:
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.
Answer:
Following are the answer to this question:
Step-by-step explanation:
In the given-question, the information is missing. first, we declare the missing information, after that we define its solution:
Missing information:
plotting the Points (2, 0), (2, 4), (2, 1), and (2, -1).
solution:
please find the attachment.
In the given attachment file, all the points lie within the same line, which indicates its points, and the set may be interpreted throughout the form of (2,y), or even the points may also be placed on the line x=2.
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