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Solve the equation on the interval [0,2π) .

6sin^2x+3sinx=0

 

What are the correct solutions?

 Apr 8, 2020
 #1
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Solve the equation on the interval \([0,2\pi)\) .
\(6\sin^2(x)+3\sin(x)=0\)


What are the correct solutions?

 

\(\begin{array}{|lrcll|} \hline &\mathbf{6\sin^2(x)+3\sin(x)} &=& \mathbf{0} \quad &| \quad : 3 \\\\ &2\sin^2(x)+ \sin(x) &=& 0 \\ &\underbrace{\sin(x)}_{=0}\Big( \underbrace{2\sin(x)+ 1}_{=0} \Big) &=& 0 \\ \\ \hline 1) & \sin(x) &=& 0 \\ & x &=& \arcsin(0) + \pi n \qquad n \in \mathbb{Z} \\ & x &=& 0 + \pi n \\ & \mathbf{x} &=& \mathbf{\pi n} \qquad n \in \mathbb{Z} \\ \hline 2) & 2\sin(x)+ 1 &=& 0 \\ & \sin(x) &=& -\dfrac{1}{2} \\ & x &=& \arcsin\left(-\dfrac{1}{2}\right) + 2\pi n \qquad n \in \mathbb{Z} \\ & \mathbf{x} &=& \mathbf{-\dfrac{\pi}{6} + 2\pi n} \qquad n \in \mathbb{Z} \\ \hline 3) & \sin(x) &=& -\dfrac{1}{2} \quad & | \quad \sin(x) = \sin(\pi-x) \\ & \sin(\pi-x) &=& -\dfrac{1}{2} \\ & \pi-x &=& \arcsin\left(-\dfrac{1}{2}\right) + 2\pi n \qquad n \in \mathbb{Z} \\ & \pi-x &=& -\dfrac{\pi}{6} + 2\pi n \qquad n \in \mathbb{Z} \\ & x &=& \pi + \dfrac{\pi}{6} + 2\pi m \qquad m \in \mathbb{Z} \\ & \mathbf{x} &=& \mathbf{\dfrac{7\pi}{6}} + 2\pi m \qquad m \in \mathbb{Z} \\ \hline \end{array}\)

 

On the interval \([0,2\pi)\)

solutions:
\(\begin{array}{|lll|} \hline x &= 0\\\\ x &= \pi\\\\ x &= \dfrac{7 \pi}{6}\\\\ x &= \dfrac{11 \pi}{6} \\ \hline \end{array}\)

 

laugh

 Apr 8, 2020

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