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2 years ago

What is the solution to the trigonometric inequality 2sin(x)+3>sin^2(x) over the interval 0<=x<=2pi radians?

Mathematics

1 answer:

devlian [24]2 years ago

7 0

2 sin(x) + 3 > sin²(x)

can be rewritten as

sin²(x) - 2 sin(x) - 3 < 0

(sin(x) - 3) (sin(x) + 1) < 0

The left side is equal to zero when either

sin(x) - 3 = 0 or sin(x) + 1 = 0

sin(x) = 3 or sin(x) = -1

The first case can never happen for real x, since |sin(x)| ≤ 1. From the second case, we get one zero in the interval 0 ≤ x ≤ 2π at x = 3π/2.

Now, split up the solution domain into two intervals, and check the sign of the left side of the inequality at some point in the interval. If the inequality holds up, that interval will belong to the solution set for the inequality.

• From 0 ≤ x < 3π/2, take x = 0. Then

sin²(0) - 2 sin(0) - 3 = 0² - 0 - 3 < 0

is true.

• From 3π/2 < x ≤ 2π, take x = 2π. Then

sin²(2π) - 2 sin(2π) - 3 = 0² - 0 - 3 = -3 < 0

is also true.

This means both intervals 0 ≤ x < 3π/2 and 3π/2 < x ≤ 2π satisfy the inequality.

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2 years ago

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Answer:

Therefore the circumference of the circle is $=\frac{20\pi}{4+\pi}$

Step-by-step explanation:

Let the side of the square be s

and the radius of the circle be r

The perimeter of the square is = 4s

The circumference of the circle is =2πr

Given that the length of the wire is 20 cm.

According to the problem,

4s + 2πr =20

⇒2s+πr =10

$\Rightarrow s=\frac{10-\pi r}{2}$

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The area of the square is = s²

A represent the total area of the square and circle.

A=πr²+s²

Putting the value of s

$A=\pi r^2+ (\frac{10-\pi r}{2})^2$

$\Rightarrow A= \pi r^2+(\frac{10}{2})^2-2.\frac{10}{2}.\frac{\pi r}{2}+ (\frac{\pi r}{2})^2$

$\Rightarrow A=\pi r^2 +25-5 \pi r +\frac{\pi^2r^2}{4}$

$\Rightarrow A=\pi r^2\frac{4+\pi}{4} -5\pi r +25$

For maximum or minimum $\frac{dA}{dr}=0$

Differentiating with respect to r

$\frac{dA}{dr}= \frac{2\pi r(4+\pi)}{4} -5\pi$

Again differentiating with respect to r

$\frac{d^2A}{dr^2}=\frac{2\pi (4+\pi)}{4}$ > 0

For maximum or minimum

$\frac{dA}{dr}=0$

$\Rightarrow \frac{2\pi r(4+\pi)}{4} -5\pi=0$

$\Rightarrow r = \frac{10\pi }{\pi(4+\pi)}$

$\Rightarrow r=\frac{10}{4+\pi}$

$\frac{d^2A}{dr^2}|_{ r=\frac{10}{4+\pi}}=\frac{2\pi (4+\pi)}{4}>0$

Therefore at $r=\frac{10}{4+\pi}$ , A is minimum.

Therefore the circumference of the circle is

$=2 \pi \frac{10}{4+\pi}$

$=\frac{20\pi}{4+\pi}$

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3 years ago