Question

Sin^2x+4cos^2x+5sinxcosx=0 решите пожалуйста

Answer 1

Ответ:

$sin^2x+4cos^2x+5sinx\cdot cosx=0\ \Big|:cos^2x\ne 0\\\\tg^2x+5tgx+4=0\ \ \ \Rightarrow \ \ \ tgx=-1\ \ ili\ \ tgx=-4\ \ (teorema\ Vieta)\\\\a)\ \ tgx=-1\ \ ,\ \ x=-\dfrac{\pi }{4}+\pi n\ \ ,\ n\in Z\\\\b)\ \ tgx=-4\ \ ,\ \ \ x=-arctg4+\pi k\ \ ,\ k\in Z\\\\Otvet:\ \ x=-\dfrac{\pi }{4}+\pi n\ \ ,\ \ x=-arctg4+\pi k\ \ ,\ n,k\in Z\ .$

Answer 2

To solve the equation sin^2x + 4cos^2x + 5sinxcosx = 0, we can use trigonometric identities and algebraic manipulation. Let's break it down step by step:

Step 1: Simplify the equation using trigonometric identities

We can rewrite sin^2x as (1 - cos^2x) using the identity sin^2x + cos^2x = 1. Substituting this into the equation, we get:

(1 - cos^2x) + 4cos^2x + 5sinxcosx = 0

Simplifying further, we have:

1 + 3cos^2x + 5sinxcosx = 0

Step 2: Rearrange the equation

Let's rearrange the equation to isolate the terms involving cosine:

3cos^2x + 5sinxcosx = -1

Step 3: Solve for cosx

We can use the identity sin2x = 2sinxcosx to rewrite the equation:

3cos^2x + 5(2sinxcosx) = -1

Simplifying further:

3cos^2x + 10sinxcosx = -1

Now, let's use the identity cos^2x = 1 - sin^2x to substitute for cos^2x:

3(1 - sin^2x) + 10sinxcosx = -1

Simplifying again:

3 - 3sin^2x + 10sinxcosx = -1

Step 4: Solve the quadratic equation

Rearranging the terms, we have:

3sin^2x - 10sinxcosx + 4 = 0

This is a quadratic equation in terms of sinx. We can solve it by factoring or using the quadratic formula. However, it seems that there might be a mistake in the original equation or the steps taken to simplify it. The equation sin^2x + 4cos^2x + 5sinxcosx = 0 does not seem to lead to a quadratic equation in terms of sinx.

Please double-check the original equation or provide additional information if there was an error in the equation or the steps taken to simplify it.