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Mathematics 14 Online
OpenStudy (anonymous):

cos^(2)x = 0 ???

OpenStudy (anonymous):

Well, I know that cos90=0 so cos^2x=cos90 can you solve for x now?

OpenStudy (jdoe0001):

\(\bf cos^2(x)=0\implies cos(x)=\sqrt{0}\implies cos(x)=0\\ \quad \\ cos^{-1}[cos(x)]=cos^{-1}(0)\implies x=cos^{-1}(0)\)

OpenStudy (anonymous):

I made it even harder, my bad.

OpenStudy (anonymous):

then cos^(2)x=0 is the same as cosx=0 or is bad?

OpenStudy (jdoe0001):

well, all I did is take square root to both sides, to isolate the cosine part, so \(\bf cos^2(x)=0\implies \sqrt{cos^2(x)}=\sqrt{0} \implies cos(x)=0\\ \quad \\ cos^{-1}[cos(x)]=cos^{-1}(0)\implies x=cos^{-1}(0)\)

OpenStudy (jdoe0001):

\(\bf 0 = 0^2\qquad thus\qquad \sqrt{0}\implies \sqrt{0^2}=0\)

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