Question

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

Answer 1

Ответ:

x=0,5·(-1)ⁿ⁺¹·arcsin(1/3)+nπ, n∈Z

Пошаговое объяснение:

3 sin²2x+10sin2x+3=0

sin2x=t, |t|≤1

3 t²+10t+3=0

D=100-36=64=8²

t₁=(-10-8)/6=-3<-1

t₂=(-10+8)/6=-1/3

sin2x=-1/3

2x=(-1)ⁿ·arcsin(-1/3)+nπ

x=0,5·(-1)ⁿ⁺¹·arcsin(1/3)+nπ, n∈Z

Answer 2

To solve the equation $3\sin^2(2x) + 10\sin(2x) + 3 = 0$, we can treat it as a quadratic equation in terms of $\sin(2x)$. Let $y = \sin(2x)$, then the equation becomes:

$3y^2 + 10y + 3 = 0$

Now, we can solve this quadratic equation for $y$. To do that, we can either factorize it or use the quadratic formula. Let's use the quadratic formula:

The quadratic formula states that for an equation of the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

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

In our case, $a = 3$, $b = 10$, and $c = 3$. Now, plug these values into the formula:

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

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

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

$y = \frac{-10 \pm 8}{6}$

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

1. $y = \frac{-10 + 8}{6} = \frac{-2}{6} = -\frac{1}{3}$
2. $y = \frac{-10 - 8}{6} = \frac{-18}{6} = -3$

Now, remember that we set $y = \sin(2x)$. So, we have:

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

However, $\sin(2x)$ is bounded between -1 and 1. So, the second solution $\sin(2x) = -3$ is not possible.

We need to find the solutions for $2x$ from the first equation $\sin(2x) = -\frac{1}{3}$:

$2x = \arcsin\left(-\frac{1}{3}\right) + 2n\pi$ or $2x = \pi - \arcsin\left(-\frac{1}{3}\right) + 2n\pi$

where $n$ is an integer.

Now, we can solve for $x$ by dividing both sides by 2:

$x = \frac{\arcsin\left(-\frac{1}{3}\right) + 2n\pi}{2}$ or $x = \frac{\pi - \arcsin\left(-\frac{1}{3}\right) + 2n\pi}{2}$

Remember that $n$ can be any integer, positive, negative, or zero. Plug in different values of $n$ to get different solutions for $x$.