Solve 3 cos^2(sigma) + 2 sin(sigma) = 1

ab5490

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Mar 28, 2007
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3cos²σ+2sinσ=1 find σ in interval 0 ≤ σ 360 to the nearest degree
 
\(\displaystyle \L 3\cos^2{x} + 2\sin{x} = 1\)

\(\displaystyle \L 3(1 - \sin^2{x}) + 2\sin{x} = 1\)

\(\displaystyle \L 0 = 3\sin^2{x} - 2\sin{x} - 2\)

\(\displaystyle \L \sin{x} = \frac{1 \pm \sqrt{7}}{3}\)

since \(\displaystyle \L \frac{1 + \sqrt{7}}{3} > 1\), only \(\displaystyle \L \sin{x} = \frac{1 - \sqrt{7}}{3}\) can be solved for x.

\(\displaystyle \L x = 360 + arcsin\left(\frac{1 - \sqrt{7}}{3}\right) \approx 327^o\)

\(\displaystyle \L x = 180 - arcsin\left(\frac{1 - \sqrt{7}}{3}\right) \approx 213^o\)
 
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