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What values for theta (0<=theta<=2pi) satify the equation?

 

2 sin theta cos theta + cos theta = 0

 Jun 6, 2019

Best Answer 

 #1
avatar+8756 
+3

\(2\sin\theta\cos\theta+\cos\theta\ =\ 0\)

                                                     Factor   cos θ   out of both terms.

\((\cos\theta)(2\sin\theta+1)\ =\ 0\)

                                                     Set each factor equal to  0  and solve for  θ .

                                                     And remember that we only want solutions for  θ  in the interval  [0, 2π]

 

\(\begin{array}{} \cos\theta\ =\ 0&\qquad\text{or}\qquad&2\sin\theta+1\ =\ 0\\~\\ \theta\ =\ \frac{\pi}{2}&&2\sin\theta\ =\ -1\\~\\ \theta\ =\ \frac{3\pi}{2}&&\sin\theta\ =\ -\frac{1}{2}\\~\\ &&\theta\ =\ \frac{7\pi}{6}\\~\\ &&\theta\ =\ \frac{11\pi}{6} \end{array}\)

 

So the values for  θ  in the interval  [0, 2π]  that satisfy the equation are:

 

\(\frac{\pi}{2},\ \frac{3\pi}{2},\ \frac{7\pi}{6},\ \frac{11\pi}{6}\)_

 Jun 6, 2019
 #1
avatar+8756 
+3
Best Answer

\(2\sin\theta\cos\theta+\cos\theta\ =\ 0\)

                                                     Factor   cos θ   out of both terms.

\((\cos\theta)(2\sin\theta+1)\ =\ 0\)

                                                     Set each factor equal to  0  and solve for  θ .

                                                     And remember that we only want solutions for  θ  in the interval  [0, 2π]

 

\(\begin{array}{} \cos\theta\ =\ 0&\qquad\text{or}\qquad&2\sin\theta+1\ =\ 0\\~\\ \theta\ =\ \frac{\pi}{2}&&2\sin\theta\ =\ -1\\~\\ \theta\ =\ \frac{3\pi}{2}&&\sin\theta\ =\ -\frac{1}{2}\\~\\ &&\theta\ =\ \frac{7\pi}{6}\\~\\ &&\theta\ =\ \frac{11\pi}{6} \end{array}\)

 

So the values for  θ  in the interval  [0, 2π]  that satisfy the equation are:

 

\(\frac{\pi}{2},\ \frac{3\pi}{2},\ \frac{7\pi}{6},\ \frac{11\pi}{6}\)_

hectictar Jun 6, 2019

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