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3 years ago

Find all solutions to the equation:

Mathematics

1 answer:

crimeas [40]3 years ago

7 0

Let z = sin(x). This means z^2 = (sin(x))^2 = sin^2(x). This allows us to go from the equation you're given to this equation: 7z^2 - 14z + 2 = -5

That turns into 7z^2 - 14z + 7 = 0 after adding 5 to both sides. Use the quadratic formula to solve for z. The only solution is z = 1 (see attached image). Since we made z = sin(x), this means sin(x) = 1. All solutions to this equation will be in the form x = (pi/2) + 2pi*n, which is the radian form of the solution set. If you need the degree form, then it would be x = 90 + 360*n

The 2pi*n (or 360*n) part ensures we get every angle coterminal to pi/2 radians (90 degrees), which captures the entire solution set.

Note: The variable n can be any integer.

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3 years ago

-62 + 18y = 264<br> -12x - 36y = 130<br><br><br> What solution is this

sergeinik [125]

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2 years ago

Sketch the equilibrium solutions for the following DE and use them to determine the behavior of the solutions.

GREYUIT [131]

Answer:

$y=\dfrac{1}{1-Ke^{-t}}$

Step-by-step explanation:

Given

The given equation is a differential equation

$\dfrac{dy}{dt}=y-y^2$

$\dfrac{dy}{dt}=-(y^2-y)$

By separating variable

⇒ $\dfrac{dy}{(y^2-y)}=-t$

$\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-dt$

Now by taking integration both side

$\int\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-\int dt$

⇒ $\ln (y-1)-\ln y=-t+C$

Where C is the constant

$\ln \dfrac{y-1}{y}=-t+C$

$\dfrac{y-1}{y}=e^{-t+c}$

$\dfrac{y-1}{y}=Ke^{-t}$

$y=\dfrac{1}{1-Ke^{-t}}$

from above equation we can say that

When t will increases in positive direction then $e^{-t}$ will decreases it means that ${1-Ke^{-t}}$ will increases, so y will decreases. Similarly in the case of negative t.