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Find the number of solutions of \sin ^{2} x+\cos x=0 \text { in }[-2 \pi, 2 \pi]

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

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\\\sin ^{2} x+\cos x=0\\ 1- \cos^2 x + \cos x = 0\\ \cos^2 x - \cos x -1= 0\\

\\\mathrm{Let}\cos \left(x\right)=u\\u^2-u-1=0\\u=\frac{1+\sqrt{5}}{2},\:u=\frac{1-\sqrt{5}}{2}\\\cos \left(x\right)=\frac{1+\sqrt{5}}{2},\:\cos \left(x\right)=\frac{1-\sqrt{5}}{2}\\\\\cos \left(x\right)\mathrm{\:can't\:be\:greater\:than\:one\:for\:real\:solutions,\,\,so\,\,first\,\,value\,\,is\,\,rejected}\\\\\cos \left(x\right)=\frac{1-\sqrt{5}}{2}\\\\\text {From the graph of y=cos(x), we know that x can attain two values in }[0,2 \pi] \text { and two more values in }[-2 \pi, 0) .\\ \text {Thus, there are four solutions. }

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