Question

2 sin квадрат x-cos x=1

Answer 1

Ответ:

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Answer 2

The equation you provided is: $2 \sin^2(x) - \cos(x) = 1.$

We can solve this equation for $x$. First, let's rearrange the equation to isolate the trigonometric terms on one side:

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

Now, we need to use trigonometric identities to simplify the equation. We'll use the Pythagorean identity for sine and cosine:

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

Rearranging this identity, we get:

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

Now we can substitute this into our equation:

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

Simplify this equation:

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

Rearrange and combine terms:

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

Divide the entire equation by -1 to make the leading coefficient of the quadratic term positive:

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

Now, this is a quadratic equation in terms of $\cos(x)$. We can solve it by factoring or using the quadratic formula:

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

Let $u = \cos(x)$, then the equation becomes:

$2u^2 + u - 1 = 0.$

We can factor this equation:

$(2u - 1)(u + 1) = 0.$

This gives us two possible values for $u$:

1. $2u - 1 = 0$ --> $u = \frac{1}{2}$
2. $u + 1 = 0$ --> $u = -1$

Now, remember that $u = \cos(x)$, so:

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

We can solve these equations for $x$:

1. For $\cos(x) = \frac{1}{2}$, there are two solutions: $x = \frac{\pi}{3} + 2\pi n$ and $x = \frac{5\pi}{3} + 2\pi n$, where $n$ is an integer.

2. For $\cos(x) = -1$, there is one solution: $x = \pi + 2\pi n$, where $n$ is an integer.

So, the solutions for the original equation are:

$x = \frac{\pi}{3} + 2\pi n, \quad x = \frac{5\pi}{3} + 2\pi n, \quad x = \pi + 2\pi n,$ where $n$ is an integer.