Question

4 sin^2(2x)- 8 sin (2x) +3=0​

Answer 1

Все такие адания решаются однотипно. Путем замены, мы получаем обычное квадратное уравнение.

Answer 2

To solve the quadratic equation $4\sin^2(2x) - 8\sin(2x) + 3 = 0$, you can use a substitution to make it easier to work with. Let's use $u = \sin(2x)$. Now the equation becomes:

$4u^2 - 8u + 3 = 0$

This is a quadratic equation in $u$. To solve it, you can use the quadratic formula:

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

In this case, $a = 4$, $b = -8$, and $c = 3$. Plug these values into the formula:

$u = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(4)(3)}}{2(4)}$

Simplify:

$u = \frac{8 \pm \sqrt{64 - 48}}{8}$

$u = \frac{8 \pm \sqrt{16}}{8}$

Now, take the square root of 16:

$u = \frac{8 \pm 4}{8}$

Now, you have two possible solutions for $u$:

1. $u_1 = \frac{8 + 4}{8} = \frac{12}{8} = \frac{3}{2}$
2. $u_2 = \frac{8 - 4}{8} = \frac{4}{8} = \frac{1}{2}$

Now, remember that $u = \sin(2x)$. So you have:

1. $\sin(2x) = \frac{3}{2}$: This equation has no solutions because the sine function's range is between -1 and 1, and $\frac{3}{2}$ is outside that range.

2. $\sin(2x) = \frac{1}{2}$: This equation can be solved. To find the solutions, use the inverse sine function:

$\sin^{-1}\left(\frac{1}{2}\right) = 30^\circ = \frac{\pi}{6}$

So, $2x = \frac{\pi}{6} + 2\pi n$ and $2x = \frac{5\pi}{6} + 2\pi n$ for integer values of $n$.

Now, divide both sides by 2 to solve for $x$:

$x = \frac{\pi}{12} + \pi n$ and $x = \frac{5\pi}{12} + \pi n$

These are the solutions for the equation $4\sin^2(2x) - 8\sin(2x) + 3 = 0$.