sin^2(x) + cos(x) + 1 = 0 In the interval [0, 2π)

sin^2(x) + cos(x) + 1 = 0 In the interval [0, 2π)

$sin^2(x)+cos(x)+1=0$

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

$1-cos^2(x)+cos(x)+1=0$

$cos^2(x)-cos(x)-2=0$

cos(x)=a

$a^2-1a-2=0$

      p        q

$a=-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}$ 

$a=\frac{1}{2}\pm\sqrt{(\frac{1}{2})^2+2}$

$a=\frac{1}{2}\pm1.5$

$a_1=2\\cos(x)=2\\deleted$

$a_2=\frac{1}{2}-1.5\\cos(x)=-1\\x=arc\ cos (-1)$

$\large x=\pi\\or\\x=180°$

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