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2. Determine the general solution of \( 2 \sin 3 x=-\cos 3 x \)

Ask by Salinas Ortiz. in South Africa
Mar 14,2025

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The general solution is \(x = \frac{\arctan(-\frac{1}{2})}{3} + \frac{k\pi}{3}\), where \(k\) is any integer.

Solución

Solve the equation \( 2 \sin(3x)=-\cos(3x) \). Solve the equation by following steps: - step0: Solve for \(x\): \(2\sin\left(3x\right)=-\cos\left(3x\right)\) - step1: Divide both sides: \(\frac{2\sin\left(3x\right)}{\cos\left(3x\right)}=-1\) - step2: Divide the terms: \(2\tan\left(3x\right)=-1\) - step3: Multiply both sides: \(2\tan\left(3x\right)\times \frac{1}{2}=-\frac{1}{2}\) - step4: Calculate: \(\tan\left(3x\right)=-\frac{1}{2}\) - step5: Use the inverse trigonometric function: \(3x=\arctan\left(-\frac{1}{2}\right)\) - step6: Add the period: \(3x=\arctan\left(-\frac{1}{2}\right)+k\pi ,k \in \mathbb{Z}\) - step7: Solve the equation: \(x=\frac{\arctan\left(-\frac{1}{2}\right)}{3}+\frac{k\pi }{3},k \in \mathbb{Z}\) The general solution of the equation \(2 \sin 3x = -\cos 3x\) is given by \(x = \frac{\arctan(-\frac{1}{2})}{3} + \frac{k\pi}{3}\), where \(k\) is an integer.

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Beyond the Answer

To solve the equation \( 2 \sin 3x = -\cos 3x \), start by rearranging it to: \[ 2 \sin 3x + \cos 3x = 0. \] This can be interpreted as: \[ \cos 3x = -2 \sin 3x. \] Dividing both sides by \( \cos 3x \) (where \( \cos 3x \neq 0 \)), we can use the tangent function: \[ \tan 3x = -\frac{1}{2}. \] Now, the general solution for \( 3x \) is: \[ 3x = \tan^{-1}(-\frac{1}{2}) + n\pi, \quad n \in \mathbb{Z}. \] Next, divide by 3 to solve for \( x \): \[ x = \frac{1}{3} \tan^{-1}(-\frac{1}{2}) + \frac{n\pi}{3}, \quad n \in \mathbb{Z}. \] This gives the general solution for \( x \).

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