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Answer 1 · 2017-04-26T17:05:36

${50.1}^{o}, {154.3}^{o}, {205.7}^{o}, {309.0}^{o}$ $50.1^o, 154.3^o, 205.7^o, 309.0^o$

Start by completing the square as this expression cannot be factorised.

$3 {cos}^{2} x + cos x - 1 = 0$ $3cos^2x+cosx-1=0$

${cos}^{2} x + \frac{1}{3} cos x - \frac{1}{3} = 0$ $cos^2x+1/3cosx-1/3=0$

${(cos x - \frac{1}{6})}^{2} - \frac{1}{36} - \frac{1}{3} = 0$ $(cosx-1/6)^2-1/36-1/3=0$

${(cos x - \frac{1}{6})}^{2} = \frac{13}{36}$ $(cosx-1/6)^2 = 13/36$

$cos x - \frac{1}{6} = \pm \sqrt{\frac{13}{36}}$ $cosx-1/6 = +-sqrt(13/36)$

$cos x = \frac{1}{6} \pm \frac{\sqrt{13}}{6}$ $cosx=1/6 +-sqrt(13)/6$

$cos x = \frac{1}{6} (1 \pm \sqrt{13})$ $cosx=1/6(1 +-sqrt(13))$

Now we look at a graph to see where the solutions will lie.

Geogebra

The graph shows solutions in radians. Solutions in degrees are as follows

${50.1}^{o}, {154.3}^{o}, {205.7}^{o}, {309.0}^{o}$ $50.1^o, 154.3^o, 205.7^o, 309.0^o$