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avatar+8079 

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

 

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

 

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

 

\(a=\frac{1}{2}\pm1.5\)

 

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

 

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

 

\(\large x=\pi\\or\\x=180°\)

 

laugh  !

 Jun 1, 2017
 #1
avatar+8079 
0

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

 

The answer is above.

 

laugh  !

 Jun 1, 2017

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