Question

Sin^2x + 3cos^2x - 2sin2x = 0

Answer 1

Ответ:

1)sin2x=2sinxcosx

2)3cos2x=3cos^2x-3sin^2x

3)2sin2x=4sinxcosx

4) 2sinxcosx+  3cos^2x-3sin^2x-  4sinxcosx= -2sinxcosx+ 3cos^2x-3sin^2x

т.к это однородное уравнение 2 степени значит делим на cos^2x

5)-2tgx+3-3tg^2x=0

6)пусть tgx=a

-2а+3-3а^2=0

ну а дальше через дискрименант  

а1=(2+корень из 40)/6          а2=(2+)/6

7)x=arctg(a1)+ПK,  x=arctg(a2)+Пk

Объяснение:

вроде так)

Answer 2

Ответ:

1)sin2x=2sinxcosx

2)3cos2x=3cos^2x-3sin^2x

3)2sin2x=4sinxcosx

4) 2sinxcosx+ 3cos^2x-3sin^2x- 4sinxcosx= -2sinxcosx+ 3cos^2x-3sin^2x

т.к это однородное уравнение 2 степени значит делим на cos^2x

5)-2tgx+3-3tg^2x=0

6)пусть tgx=a

-2а+3-3а^2=0

ну а дальше через дискрименант

а1=(2+корень из 40)/6 а2=(2+)/6

7)x=arctg(a1)+ПK, x=arctg(a2)+Пk

Answer 3

To solve the trigonometric equation $\sin^2(x) + 3\cos^2(x) - 2\sin(2x) = 0$, we can use trigonometric identities and algebraic manipulation.

First, let's use the double angle identity for sine: $\sin(2x) = 2\sin(x)\cos(x)$. Now, we can rewrite the equation as:

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

Now, let's simplify this equation:

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

Next, we can use the Pythagorean identity for sine: $\sin^2(x) + \cos^2(x) = 1$. So, $\cos^2(x) = 1 - \sin^2(x)$. We can substitute this into the equation:

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

Now, distribute the 3 on the second term:

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

Combine like terms:

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

Now, factor out a 2 from the second and third terms:

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

Now, we can notice that $\sin^2(x) + 2\sin(x)\cos(x)$ is the double angle identity for sine again: $\sin(2x)$. So, we have:

$3 - 2\sin(2x) = 0$

Now, isolate $\sin(2x)$:

$2\sin(2x) = 3$

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

However, the sine function has a maximum value of 1, and $\frac{3}{2}$ is greater than 1, which means there are no real solutions for $\sin(2x)$ in this equation.

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