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aleksandrvk [35]
2 years ago
11

How many solutions does the equation sin(5x) = 1/2 have on the interval (0, 2PI]

Mathematics
1 answer:
Darina [25.2K]2 years ago
4 0

Answer:

Step-by-step explanation:

Given the equation

Sin(5x) = ½

5x = arcSin(½)

5x = 30°

Then,

The general formula for sin is

5θ = n180 + (-1)ⁿθ

Divide through by 5

θ = n•36 + (-1)ⁿ30/5

θ = 36n + (-1)ⁿ6

The range of the solution is

0<θ<2π I.e 0<θ<360

First solution

When n = 0

θ = 36n + (-1)ⁿθ

θ = 0×36 + (-1)^0×6

θ = 6°

When n = 1

θ = 36n + (-1)ⁿ6

θ = 36-6

θ = 30°

When n = 2

θ = 36n + (-1)ⁿ6

θ = 36×2 + 6

θ = 78°

When n =3

θ = 36n + (-1)ⁿ6

θ = 36×3 - 6

θ = 102°

When n=4

θ = 36n + (-1)ⁿ6

θ = 36×4 + 6

θ = 150

When n =5

θ = 36n + (-1)ⁿ6

θ = 36×5 - 6

θ = 174°

When n = 6

θ = 36n+ (-1)ⁿ6

θ = 36×6 + 6

θ = 222°

When n = 7

θ = 36n + (-1)ⁿ6

θ = 36×7 - 6

θ = 246°

When n =8

θ = 36n + (-1)ⁿ6

θ = 36×8 + 6

θ = 294°

When n =9

θ = 36n + (-1)ⁿ6

θ = 36×9 - 6

θ = 318°

When n =10

θ = 36n + (-1)ⁿ6

θ = 36×10 + 6

θ = 366°

When n = 10 is out of range of θ

Then, the solution is from n =0 to n=9

So the equation have 10 solutions in the range 0<θ<2π

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Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into
Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = (\frac{60\pi }{\pi+4}) in

b)the length of the wire for the square = (\frac{240}{\pi+4}) in

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

         b=\frac{y}{4}

Area of the square which is L² can now be said to be;

A_S=(\frac{y}{4})^2 = \frac{y^2}{16}

On the otherhand; let the radius (r) of the  circle be;

2πr = 60-y

r = \frac{60-y}{2\pi }

Area of the circle which is πr² can now be;

A_C= \pi (\frac{60-y}{2\pi } )^2

     =( \frac{60-y}{4\pi } )^2

Total Area (A);

A = A_S+A_C

   = \frac{y^2}{16} +(\frac{60-y}{4\pi } )^2

For the smallest possible area; \frac{dA}{dy}=0

∴ \frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0

If we divide through with (2) and each entity move to the opposite side; we have:

\frac{y}{18}=\frac{(60-y)}{2\pi}

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

y= \frac{240}{\pi+4}

∴ since y= \frac{240}{\pi+4}, we can determine for the length of the circle ;

60-y can now be;

= 60-\frac{240}{\pi+4}

= \frac{(\pi+4)*60-240}{\pi+40}

= \frac{60\pi+240-240}{\pi+4}

= (\frac{60\pi}{\pi+4})in

also, the length of wire for the square  (y) ; y= (\frac{240}{\pi+4})in

The smallest possible area (A) = \frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0
2 years ago
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