3cos²σ+2sinσ=1 find σ in interval 0 ≤ σ 360 to the nearest degree
A ab5490 New member Joined Mar 28, 2007 Messages 3 Mar 28, 2007 #1 3cos²σ+2sinσ=1 find σ in interval 0 ≤ σ 360 to the nearest degree
skeeter Elite Member Joined Dec 15, 2005 Messages 3,216 Mar 29, 2007 #2 \(\displaystyle \L 3\cos^2{x} + 2\sin{x} = 1\) \(\displaystyle \L 3(1 - \sin^2{x}) + 2\sin{x} = 1\) \(\displaystyle \L 0 = 3\sin^2{x} - 2\sin{x} - 2\) \(\displaystyle \L \sin{x} = \frac{1 \pm \sqrt{7}}{3}\) since \(\displaystyle \L \frac{1 + \sqrt{7}}{3} > 1\), only \(\displaystyle \L \sin{x} = \frac{1 - \sqrt{7}}{3}\) can be solved for x. \(\displaystyle \L x = 360 + arcsin\left(\frac{1 - \sqrt{7}}{3}\right) \approx 327^o\) \(\displaystyle \L x = 180 - arcsin\left(\frac{1 - \sqrt{7}}{3}\right) \approx 213^o\)
\(\displaystyle \L 3\cos^2{x} + 2\sin{x} = 1\) \(\displaystyle \L 3(1 - \sin^2{x}) + 2\sin{x} = 1\) \(\displaystyle \L 0 = 3\sin^2{x} - 2\sin{x} - 2\) \(\displaystyle \L \sin{x} = \frac{1 \pm \sqrt{7}}{3}\) since \(\displaystyle \L \frac{1 + \sqrt{7}}{3} > 1\), only \(\displaystyle \L \sin{x} = \frac{1 - \sqrt{7}}{3}\) can be solved for x. \(\displaystyle \L x = 360 + arcsin\left(\frac{1 - \sqrt{7}}{3}\right) \approx 327^o\) \(\displaystyle \L x = 180 - arcsin\left(\frac{1 - \sqrt{7}}{3}\right) \approx 213^o\)