Question

A)2sin^2x-2√2cosx+1=0 б)корни’принадлежащие отрезку 5p/2;4p

Answer 1

Ответ:

а) $x=\pm\frac{\pi}{4}+2\pi*n,\,\,n\in Z$

б) $x=3\frac{3}{4}\pi$

Объяснение:

Решение в приложении

Answer 2

To solve the quadratic equation $2\sin^2x - 2\sqrt{2}\cos x + 1 = 0$ and find the roots that belong to the interval $\left[\frac{5\pi}{2}, 4\pi\right]$, we'll need to use some trigonometric identities.

Let's rewrite the equation in terms of trigonometric functions:

$2\sin^2x - 2\sqrt{2}\cos x + 1 = 0$

Using the identity $\sin^2x + \cos^2x = 1$, we can rewrite $\sin^2x$ as $1 - \cos^2x$:

$2(1 - \cos^2x) - 2\sqrt{2}\cos x + 1 = 0$

Simplify the equation:

$2 - 2\cos^2x - 2\sqrt{2}\cos x + 1 = 0$

Combine like terms:

$-2\cos^2x - 2\sqrt{2}\cos x + 3 = 0$

Now, let's make a substitution to simplify the equation further. Let $t = \cos x$. Then, the equation becomes:

$-2t^2 - 2\sqrt{2}t + 3 = 0$

Now, we have a standard quadratic equation in $t$. We can solve for $t$ using the quadratic formula:

For a quadratic equation of the form $at^2 + bt + c = 0$, the solutions are given by:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our case, $a = -2$, $b = -2\sqrt{2}$, and $c = 3$.

$t = \frac{-(-2\sqrt{2}) \pm \sqrt{(-2\sqrt{2})^2 - 4(-2)(3)}}{2(-2)}$

$t = \frac{2\sqrt{2} \pm \sqrt{8 + 24}}{-4}$

$t = \frac{2\sqrt{2} \pm \sqrt{32}}{-4}$

$t = \frac{2\sqrt{2} \pm 4\sqrt{2}}{-4}$

Now, we get two possible values for $t$:

1. $t = \frac{2\sqrt{2} + 4\sqrt{2}}{-4} = \frac{6\sqrt{2}}{-4} = -\frac{3\sqrt{2}}{2}$