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Вопрос от Анонимного юзера 15 февраля 2025 15:44

Sinx+√(3/2)(1–cosx) = 0,

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15 февраля 2025 15:44

To solve the equation ( \sin x + \sqrt{\frac{3}{2}}(1 - \cos x) = 0 ), let's start by rearranging the equation:

[ \sin x = -\sqrt{\frac{3}{2}}(1 - \cos x) ]

Now, we can work further with this equation. Recall that (1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)). We can replace (1 - \cos x) in the original equation:

[ \sin x = -\sqrt{\frac{3}{2}} \cdot 2 \sin^2\left(\frac{x}{2}\right) ]

This rewrites to:

[ \sin x = -\sqrt{6} \sin^2\left(\frac{x}{2}\right) ]

Next, we can use the double angle formula for sine, which states that ( \sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) ):

[ 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) = -\sqrt{6} \sin^2\left(\frac{x}{2}\right) ]

Assuming ( \sin\left(\frac{x}{2}\right) \neq 0 ), we can divide both sides by ( \sin\left(\frac{x}{2}\right) ):

[ 2 \cos\left(\frac{x}{2}\right) = -\sqrt{6} \sin\left(\frac{x}{2}\right) ]

Rearranging gives:

[ \sqrt{6} \sin\left(\frac{x}{2}\right) + 2 \cos\left(\frac{x}{2}\right) = 0 ]

Now, to find the angles, we can express this in terms of a tangent function:

[ \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = -\frac{2}{\sqrt{6}} \implies \tan\left(\frac{x}{2}\right) = -\frac{2}{\sqrt{6}} = -\frac{\sqrt{6}}{3} ]

Now, we need to find the angles corresponding to this tangent value. The tangent function is negative in the second and fourth quadrants. Therefore, we find:

[ \frac{x}{2} = \tan^{-1}\left(-\frac{\sqrt{6}}{3}\right) + k\pi, \quad k \in \mathbb{Z} ]

Now, we can find specific angles:

[ \frac{x}{2} = -\theta + k\pi ]

where ( \theta = \tan^{-1}(\frac{\sqrt{6}}{3}) ).

Thus, (x) becomes:

[ x = 2(-\theta + k\pi) = -2\theta + 2k\pi ]

To find numerical approximations, use a calculator to determine ( \theta ):

[ \theta \approx \tan^{-1}\left(\frac{\sqrt{6}}{3}\right) \approx 0.5236 \text{ radians} ]

Thus, the general solution becomes:

[ x \approx -2 \cdot 0.5236 + 2k\pi = -1.0472 + 2k\pi \quad \text{ for } k \in \mathbb{Z} ]

This can be expressed as:

[ x \approx 2k\pi - 1.0472 ]

Thus, the complete solution is:

[ x = -2\theta + 2k\pi \quad \text{where } k \in \mathbb{Z} ]

This gives all the possible (x) values satisfying the original equation.

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sinx+√(3/2)(1–cosx) = 0,
To solve the equation \( \sin x + \sqrt{\frac{3}{2}}(1 - \cos x) = 0 \), let's start by rearranging the equation: \[ \sin x = -\sqrt{\frac{3}{2}}(1 - \cos x) \] Now, we can work further with this equation. Recall that \(1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)\). We can replace \(1 - \cos x\) in the original equation: \[ \sin x = -\sqrt{\frac{3}{2}} \cdot 2 \sin^2\left(\frac{x}{2}\right) \] This rewrites to: \[ \sin x = -\sqrt{6} \sin^2\left(\frac{x}{2}\right) \] Next, we can use the double angle formula for sine, which states that \( \sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \): \[ 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) = -\sqrt{6} \sin^2\left(\frac{x}{2}\right) \] Assuming \( \sin\left(\frac{x}{2}\right) \neq 0 \), we can divide both sides by \( \sin\left(\frac{x}{2}\right) \): \[ 2 \cos\left(\frac{x}{2}\right) = -\sqrt{6} \sin\left(\frac{x}{2}\right) \] Rearranging gives: \[ \sqrt{6} \sin\left(\frac{x}{2}\right) + 2 \cos\left(\frac{x}{2}\right) = 0 \] Now, to find the angles, we can express this in terms of a tangent function: \[ \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = -\frac{2}{\sqrt{6}} \implies \tan\left(\frac{x}{2}\right) = -\frac{2}{\sqrt{6}} = -\frac{\sqrt{6}}{3} \] Now, we need to find the angles corresponding to this tangent value. The tangent function is negative in the second and fourth quadrants. Therefore, we find: \[ \frac{x}{2} = \tan^{-1}\left(-\frac{\sqrt{6}}{3}\right) + k\pi, \quad k \in \mathbb{Z} \] Now, we can find specific angles: \[ \frac{x}{2} = -\theta + k\pi \] where \( \theta = \tan^{-1}(\frac{\sqrt{6}}{3}) \). Thus, \(x\) becomes: \[ x = 2(-\theta + k\pi) = -2\theta + 2k\pi \] To find numerical approximations, use a calculator to determine \( \theta \): \[ \theta \approx \tan^{-1}\left(\frac{\sqrt{6}}{3}\right) \approx 0.5236 \text{ radians} \] Thus, the general solution becomes: \[ x \approx -2 \cdot 0.5236 + 2k\pi = -1.0472 + 2k\pi \quad \text{ for } k \in \mathbb{Z} \] This can be expressed as: \[ x \approx 2k\pi - 1.0472 \] Thus, the complete solution is: \[ x = -2\theta + 2k\pi \quad \text{where } k \in \mathbb{Z} \] This gives all the possible \(x\) values satisfying the original equation.