Question

2sin(2x+pi/3)-корень3×sinx=sin2x+корень3

Answer 1

Решение дано на фото.

Answer 2

To solve the equation $2\sin(2x+\frac{\pi}{3}) - \sqrt{3} \times \sin(x) = \sin(2x) + \sqrt{3}$, we'll follow these steps:

Step 1: Apply trigonometric identities to simplify the equation. Step 2: Use algebraic techniques to solve for $x$.

Let's get started:

Step 1: Apply trigonometric identities

First, let's simplify the left side of the equation using trigonometric identities.

We know the following identities:

1. $\sin(2x) = 2\sin(x)\cos(x)$
2. $\sin(x + y) = \sin(x)\cos(y) + \cos(x)\sin(y)$
3. $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$

Now, applying these identities:

$2\sin(2x+\frac{\pi}{3}) - \sqrt{3} \times \sin(x) = 2\left[\sin(2x)\cos(\frac{\pi}{3}) + \cos(2x)\sin(\frac{\pi}{3})\right] - \sqrt{3} \times \sin(x)$ $= 2\left[2\sin(x)\cos(x)\cos(\frac{\pi}{3}) + (1 - 2\sin^2(x))\sin(\frac{\pi}{3})\right] - \sqrt{3} \times \sin(x)$ $= 2\left[2\sin(x)\cos(x) \times \frac{1}{2} + (1 - 2\sin^2(x)) \times \frac{\sqrt{3}}{2}\right] - \sqrt{3} \times \sin(x)$ $= 2\left[\sin(x)\cos(x) + \frac{\sqrt{3}}{2} - \sqrt{3}\sin^2(x)\right] - \sqrt{3} \times \sin(x)$

Now, let's simplify the right side of the equation using the identity $\sin(2x) = 2\sin(x)\cos(x)$:

$\sin(2x) + \sqrt{3} = 2\sin(x)\cos(x) + \sqrt{3}$

Step 2: Solve for $x$

Now that we have simplified the equation, let's continue solving for $x$:

Our equation becomes:

$2\left[\sin(x)\cos(x) + \frac{\sqrt{3}}{2} - \sqrt{3}\sin^2(x)\right] - \sqrt{3} \times \sin(x) = 2\sin(x)\cos(x) + \sqrt{3}$