Question

3sin^2 2x+10sin2x+3=0

Answer 1

Решаем как квадратное уравнение
$D=100-4*3*3=100-36=64$
$sin2X_{1} = \frac{-10+8}{6} =- \frac{1}{3}$
$sin2X_{2} = \frac{-10-8}{6} =-3$ <-1 - решения нет

$sin2X=- \frac{1}{3}$
$X= \frac{(-1)^{k+1} arcsin \frac{1}{3}}{2} + \frac{ \pi k}{2}$

Answer 2

To solve the equation $3\sin^2(2x) + 10\sin(2x) + 3 = 0$, you can use the quadratic formula. Let's denote $\sin(2x)$ as $y$:

So, the equation becomes $3y^2 + 10y + 3 = 0$.

Now, you can apply the quadratic formula, which is:

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

In this case, $a = 3$, $b = 10$, and $c = 3$. Plugging these values into the formula:

$y = \frac{-10 \pm \sqrt{10^2 - 4(3)(3)}}{2(3)}$

Now, let's simplify this expression:

$y = \frac{-10 \pm \sqrt{100 - 36}}{6}$ $y = \frac{-10 \pm \sqrt{64}}{6}$ $y = \frac{-10 \pm 8}{6}$

Now, we have two possible solutions for $y$:

1. $y = \frac{-10 + 8}{6} = -\frac{2}{6} = -\frac{1}{3}$
2. $y = \frac{-10 - 8}{6} = -3$

Now, recall that $y = \sin(2x)$. So, you have two equations to solve for $2x$:

1. $\sin(2x) = -\frac{1}{3}$
2. $\sin(2x) = -3$

The first equation, $\sin(2x) = -\frac{1}{3}$, can be solved by finding the arcsin of $-1/3$, but it's important to note that the range of the arcsin function is limited to $-π/2 ≤ arcsin(y) ≤ π/2$. This means that $\sin(2x)$ can only take values in this range.

So, for $\sin(2x) = -\frac{1}{3}$, you'll find that there are no real solutions within the given range.

The second equation, $\sin(2x) = -3$, also has no real solutions because the sine function is bounded between -1 and 1.

Therefore, the original equation $3\sin^2(2x) + 10\sin(2x) + 3 = 0$ has no real solutions.