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

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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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asinus  Jun 1, 2017
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sin^2(x) + cos(x) + 1 = 0 In the interval [0, 2π)

The answer is above.

!

asinus  Jun 1, 2017

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