dwindatt8 dwindatt8

16-08-2017
Mathematics

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Solve cos x +sqr root of 2 = -cos x for x over the interval 0,2pi

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jdoe0001 jdoe0001

16-08-2017

[tex]\bf cos(x)+\sqrt{2}=-cos(x)\implies 2cos(x)+\sqrt{2}=0\implies 2cos(x)=-\sqrt{2} \\\\\\ cos(x)=-\cfrac{\sqrt{2}}{2}\implies \measuredangle x= \begin{cases} \frac{3\pi }{4}\\ \frac{5\pi }{4} \end{cases}[/tex]

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