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Mathematics 21 Online
OpenStudy (anonymous):

cosx =-1/2 find x?

OpenStudy (lgbasallote):

take the arc cosine of both sides

OpenStudy (anonymous):

What?

Parth (parthkohli):

Keep it with you :)

Parth (parthkohli):

Find the degree of the circle where \(x = \Large {-1 \over 2}\).

OpenStudy (anonymous):

is there any method to find it.... thxx for the pic

OpenStudy (lgbasallote):

here's a hint \[\cos x = 1\] \[\cos^{-1} (\cos x ) = \cos^{-1} 1\] \[x = \cos^{-1} 1\] \[x = 0\]

Parth (parthkohli):

I don't see \(\cos 0 = \Large{-1 \over 2}\)

OpenStudy (lgbasallote):

that's how you do what i said "take the arc cos of both sides"

Parth (parthkohli):

\[\cos 0 = 1 \ne {-1 \over 2} \]

OpenStudy (lgbasallote):

who said it was -1/2 o.O

Parth (parthkohli):

The question.

OpenStudy (lgbasallote):

i meant who said cos 0 = -1/2?

Parth (parthkohli):

Your solution says \(x = 0\).

OpenStudy (lgbasallote):

lol look at the start of my solution

Parth (parthkohli):

I see.

Parth (parthkohli):

You still can't evaluate \(\cos ^{-1}({-1 \over 2})\) without using a calculator/unit circle.

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