Question

3cos2x+2(sinx)2степень+4sin (x)=0

Answer 1

Cos2x=1-2sin²x
3-6sin²x+2sin²x+4sinx=0
4sin²x-4sinx-3=0
пусть sinx=t , t∈[-1;1]
тогда
4t²-4t-3=0
D=64
t₁=-1/2
t₂=3/2 не подходит по ОДЗ
sinx=-1/2
x=((-1)^n)arcsin(-π/6)+πn, n∈Z

Answer 2

To solve the trigonometric equation $3\cos^2(2x) + 2\sin^2(x) + 4\sin(x) = 0$, you can use trigonometric identities to simplify it and then find the values of $x$ that satisfy the equation.

Let's simplify the equation using trigonometric identities:

1. Use the double angle identity for cosine: $\cos(2x) = 2\cos^2(x) - 1$.

The equation becomes:

$3(2\cos^2(x) - 1) + 2\sin^2(x) + 4\sin(x) = 0$

2. Distribute and simplify:

$6\cos^2(x) - 3 + 2\sin^2(x) + 4\sin(x) = 0$

3. Combine like terms:

$6\cos^2(x) + 2\sin^2(x) + 4\sin(x) - 3 = 0$

4. Now, use the Pythagorean identity for sine and cosine: $\sin^2(x) = 1 - \cos^2(x)$.

Substitute this into the equation:

$6\cos^2(x) + 2(1 - \cos^2(x)) + 4\sin(x) - 3 = 0$

5. Simplify further:

$6\cos^2(x) + 2 - 2\cos^2(x) + 4\sin(x) - 3 = 0$

6. Combine like terms:

$4\cos^2(x) + 4\sin(x) - 1 = 0$

7. Use the trigonometric identity $\sin(x) = \sqrt{1 - \cos^2(x)}$ to express sin(x) in terms of cos(x):

$4\cos^2(x) + 4\sqrt{1 - \cos^2(x)} - 1 = 0$

8. Let $u = \cos(x)$ and rewrite the equation:

$4u^2 + 4\sqrt{1 - u^2} - 1 = 0$

Now, we have a quadratic equation in $u$. To solve for $u$, you can use the quadratic formula:

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

In this case, $a = 4$, $b = 4$, and $c = -1$. Plug these values into the formula and solve for $u$:

$u = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4}$

$u = \frac{-4 \pm \sqrt{16 + 16}}{8}$

$u = \frac{-4 \pm \sqrt{32}}{8}$

Now, simplify the square root:

$u = \frac{-4 \pm 4\sqrt{2}}{8}$

Simplify further:

$u = \frac{-1 \pm \sqrt{2}}{2}$

Now that we have the possible values of $u$, we can find corresponding values of $x$ by using the inverse cosine function:

1. $u = \frac{-1 + \sqrt{2}}{2}$