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Studentka2010 [4]
2 years ago
6

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

Mathematics
1 answer:
Fed [463]2 years ago
7 0

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.

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Equivalent expressions are expressions of equal values

The simplified expression of (3 - 4i) + (7 - 6i) is 10- 10i

The expression is given as:

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brainly.com/question/2972832

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