Question #c21c1

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Answer 1 · 2018-01-12T14:21:09

x=0°,180°,360°

${cos}^{2} x = {sin}^{2} x + 1$ $Cos^2x=sin^2x + 1$

( ${sin}^{2} x = 1 - {cos}^{2}$ $sin^2x=1 - cos^2$ )

${cos}^{2} x = 1 - {cos}^{2} + 1$ $cos^2x=1 - cos^2 + 1$

${cos}^{2} x + {cos}^{2} x = 1 + 1$ $cos^2x + cos^2x = 1+1$

$2 {cos}^{2} x = 2$ $2cos^2x = 2$

${cos}^{2} x = 1$ $cos^2x = 1$

$cos x = \pm 1$ $cosx = +-1$

This is possible when x=0°,180°,360°

Hope it helps you