Question

1) 5sin²x-2sinxcosx+cos²x=0

Answer 1

Ответ:

разделим на cosx в квадрате

$5 {tg}^{2} x - 2tgx + 1 = 0$

Замена:

$tgx = t \\ 5 {t}^{2} - 2t + 1 = 0 \\ d = 4 - 40 < 0$

Дискриминант отрицательный, значит корней нет.

Answer 2

To solve the equation $5\sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) = 0$, we can use trigonometric identities to simplify it. First, notice that $\sin^2(x) + \cos^2(x) = 1$ is the Pythagorean trigonometric identity. We can rewrite the equation as follows:

$5\sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) = 5\sin^2(x) + \cos^2(x) - 2\sin(x)\cos(x) = 1 - 2\sin(x)\cos(x) = 0$

Now, we have the equation $1 - 2\sin(x)\cos(x) = 0$. To solve for $x$, we can rearrange it:

$2\sin(x)\cos(x) = 1$

Now, we can use the double angle identity for sine to simplify it further:

$\sin(2x) = 1$

Now, we have $\sin(2x) = 1$. To solve for $x$, we need to find the angles whose sine is equal to 1. Sine is equal to 1 when the angle is $\frac{\pi}{2}$ or any angle that differs from $\frac{\pi}{2}$ by a multiple of $2\pi$. So, we can write:

$2x = \frac{\pi}{2} + 2\pi n$

where $n$ is an integer. Now, divide both sides by 2 to solve for $x$:

$x = \frac{\pi}{4} + \pi n$

So, the solutions to the equation $5\sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) = 0$ are:

$x = \frac{\pi}{4} + \pi n$

where $n$ is an integer.