Question

Cos^2x - 4sinxcosx + 3sin^2x = 0

Answer 1

$cos^2x - 4sinxcosx + 3sin^2x = 0/:cos^2x\\3tg^2x-4tgx+1=0$

Замена и метод переброски:

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, k∈Ζ

$tgx=\frac{1}{3} \\x=arctg(\frac{1}{3} )+\pi k$

, k∈Ζ

Answer 2

To solve the equation $\cos^2(x) - 4\sin(x)\cos(x) + 3\sin^2(x) = 0$, we can use trigonometric identities to simplify it and then factor it:

1. Use the trigonometric identity $\sin^2(x) + \cos^2(x) = 1$ to replace $\cos^2(x)$ with $1 - \sin^2(x)$:

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

2. Expand $(1 - \sin^2(x))$ and $3\sin^2(x)$:

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

3. Combine like terms:

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

4. Simplify the equation further:

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

5. Factor out a 2 from the first two terms:

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

6. Rewrite $2\sin^2(x)$ as $\sin^2(x) + \sin^2(x)$ and factor by grouping:

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

7. Notice that $(\sin^2(x) - 2\sin(x)\cos(x) + 1)$ can be factored as $(\sin(x) - 1)^2$:

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

8. Now, you have a quadratic equation in terms of $\sin(x)$. Set each term to zero:

$\sin(x) - 1 = 0$ (1st term)

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

Let's solve each equation separately:

For the 1st term:

$\sin(x) - 1 = 0$

$\sin(x) = 1$

This equation has solutions when $x = \frac{\pi}{2} + 2\pi n$, where $n$ is an integer.

Now, let's solve the 2nd term:

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

This equation can be factored as $(\sin(x) - \cos(x))^2 = 0$.

So, $\sin(x) - \cos(x) = 0$

$\sin(x) = \cos(x)$

This equation has solutions when $x = \frac{\pi}{4} + \pi n$, where $n$ is an integer.

So, the solutions to the original equation $\cos^2(x) - 4\sin(x)\cos(x) + 3\sin^2(x) = 0$ are:

1. $x = \frac{\pi}{2} + 2\pi n$ (where $n$