H heatly Apr 2009 102 1 May 19, 2010 #1 trig equa solve for x: 2(sinx)^2 + 7sinx-4=0 between 0 and 2pi. How would yo go about doing this one>?

trig equa solve for x: 2(sinx)^2 + 7sinx-4=0 between 0 and 2pi. How would yo go about doing this one>?

bigwave Nov 2009 717 133 Wahiawa, Hawaii May 19, 2010 #2 un FOIL heatly said: trig equa solve for x: 2(sinx)^2 + 7sinx-4=0 between 0 and 2pi. How would yo go about doing this one>? Click to expand... 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: May 19, 2010 Reactions: heatly

un FOIL heatly said: trig equa solve for x: 2(sinx)^2 + 7sinx-4=0 between 0 and 2pi. How would yo go about doing this one>? Click to expand... 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.