Solving trigonometric functions (1 Viewer)

BlueGas · Aug 23, 2015

I have trouble knowing what to do from step 4 to step 5, please take it easy on me

InteGrand · Aug 23, 2015

BlueGas said:
I have trouble knowing what to do from step 4 to step 5, please take it easy on me

$If $\cos \theta = -\frac{1}{2}$ (in our case, $\theta$ is just $\frac{\pi}{6}t$), then the acute angle $\alpha$ satisfies $\cos \alpha =\frac{1}{2}\Rightarrow \alpha = \cos ^{-1}\frac{1}{2} = \frac{\pi}{3}$. Since $\cos \theta$ is negative, draw a quadrants diagram, and in your quadrants diagram, draw angles of $\frac{\pi}{3}$ to the horizontal in quadrants 2 and 3 (since cosine in negative in and only in these quadrants). Now just read off the angles starting from the smallest positive value (positive since we are talking about time) and these are the values of $\frac{\pi}{6}t$. Then, rearrange to solve for $t$.$

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Solving trigonometric functions (1 Viewer)

BlueGas

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InteGrand

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