Find the general solutions of the following equations. \( \begin{array}{ll}\text { (a) } \sqrt{3} \cos \theta-3 \sin \theta=0 & \text { (b) } 2 \cos ^{2} \theta=\cos \theta \\ \text { (c) }(\cos \theta+\sin \theta)(1-2 \sin \theta)=0 & \text { (d) } 2 \sin ^{2} \theta-\sin \theta=1\end{array} \)
Solución de inteligencia artificial de Upstudy
Responder
Solución

¡Inicia sesión para desbloquear respuestas gratis!
Una plataforma de aprendizaje en la que confían millones de estudiantes y profesores reales.

Respondido por UpStudy AI y revisado por un tutor profesional
The Deep Dive
To solve the equations, let's tackle each of them one by one. (a) Start with the equation \(\sqrt{3} \cos \theta - 3 \sin \theta = 0\). Rearranging this, we have \(\sqrt{3} \cos \theta = 3 \sin \theta\). Dividing both sides by \(\cos \theta\) (where \(\cos \theta \neq 0\)), we get \(\sqrt{3} = 3 \tan \theta\) or \(\tan \theta = \frac{\sqrt{3}}{3}\). This occurs when \(\theta = \frac{\pi}{6} + n\pi\), where \(n\) is any integer. (b) The equation \(2 \cos^2 \theta = \cos \theta\) can be rewritten as \(2 \cos^2 \theta - \cos \theta = 0\). Factoring gives \(\cos \theta (2 \cos \theta - 1) = 0\). This results in \(\cos \theta = 0\) or \(2 \cos \theta = 1\) (i.e., \(\cos \theta = \frac{1}{2}\). The first case gives \(\theta = \frac{\pi}{2} + n\pi\) and the second case gives \(\theta = \frac{\pi}{3} + 2n\pi\) or \(\theta = \frac{5\pi}{3} + 2n\pi\). (c) For the equation \((\cos \theta + \sin \theta)(1 - 2 \sin \theta) = 0\), we solve the two factors separately. The first factor gives \(\cos \theta + \sin \theta = 0\), which can be rearranged to \(\tan \theta = -1\), yielding \(\theta = \frac{3\pi}{4} + n\pi\). The second factor gives \(1 - 2 \sin \theta = 0\) leading to \(\sin \theta = \frac{1}{2}\), or \(\theta = \frac{\pi}{6} + 2n\pi\) or \(\theta = \frac{5\pi}{6} + 2n\pi\). (d) Finally, with \(2 \sin^2 \theta - \sin \theta = 1\), we rearrange this to \(2 \sin^2 \theta - \sin \theta - 1 = 0\). Factoring yields \((2 \sin \theta + 1)(\sin \theta - 1) = 0\). This gives \(\sin \theta = -\frac{1}{2}\), or \(\theta = \frac{7\pi}{6} + 2n\pi\) and \(\theta = \frac{11\pi}{6} + 2n\pi\), and for \(\sin \theta = 1\), \(\theta = \frac{\pi}{2} + 2n\pi\). In summary, the solutions for each part are: - (a) \(\theta = \frac{\pi}{6} + n\pi\) - (b) \(\theta = \frac{\pi}{2} + n\pi, \frac{\pi}{3} + 2n\pi, \frac{5\pi}{3} + 2n\pi\) - (c) \(\theta = \frac{3\pi}{4} + n\pi, \frac{\pi}{6} + 2n\pi, \frac{5\pi}{6} + 2n\pi\) - (d) \(\theta = \frac{7\pi}{6} + 2n\pi, \frac{11\pi}{6} + 2n\pi, \frac{\pi}{2} + 2n\pi\)
