trig equa solve for x: 2(sinx)^2 + 7sinx-4=0 between 0 and 2pi. How would yo go about doing this one>?
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Originally Posted by heatly trig equa solve for x: 2(sinx)^2 + 7sinx-4=0 between 0 and 2pi. How would yo go about doing this one>? try to un FOIL it $\displaystyle (2sinx - 1)(sinx + 4) = 0$ the rest is easy... $\displaystyle 2sinx = 1 \rightarrow sinx = \frac{1}{2} \rightarrow sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\ ,\frac{5\pi}{6}$ $\displaystyle sinx = -4$ is undefined.
Last edited by bigwave; May 19th 2010 at 09:41 PM. Reason: more steps
Thank you very much,very good.
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