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klio [65]
3 years ago
6

How to solve sin3x+cos2x+2=0?

Mathematics
2 answers:
harkovskaia [24]3 years ago
3 0
It depends on which field this equation is defined. Here you have some solution for real and the beginning of calculations defined in complex field.

s344n2d4d5 [400]3 years ago
3 0
Hello,

sin (3x)=3sin(x)-4sin^3(x)
cos(2x)=1-2sin²(x)

Thus -4sin ^3(x)-2sin²(x)+3sin(x)+3=0
Lest's assume y=sin (x)

4y^3+2y²-3y-3=0
==>(y-1)(4y²+6y+3)=0
Only one real solution: sin(x)=1==>x=π/2+2kπ
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Answer:

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Step-by-step explanation:

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a) Observe that the variable x='<em>classes taken in one semester' </em>can take the values 0,1,2,...,n.

Then the variable x is discrete

b) Observe that the variable x='<em>annual rainfall in a city'</em> can take the values 2in, 1.6in, 5.1 in, 0.1in

Then, the variable x can be take a infinite number of values between two number. So x isn't a discrete variable.

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d) The variable x='<em>patients treated at an emergency room in a day</em>' can take the values 1,2,3,...,n. And never will be a decimal number because There cannot be a personal decimal number. Therefore, x is a discrete variable.

7 0
3 years ago
Answer these 2 questions please
Harlamova29_29 [7]
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3 years ago
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Solution-

Answer is <u>same,</u> <u>same.</u>

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7 0
3 years ago
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