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cosx =-1/2 find x?
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take the arc cosine of both sides
What?
Keep it with you :)
Find the degree of the circle where \(x = \Large {-1 \over 2}\).
is there any method to find it.... thxx for the pic
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here's a hint \[\cos x = 1\] \[\cos^{-1} (\cos x ) = \cos^{-1} 1\] \[x = \cos^{-1} 1\] \[x = 0\]
I don't see \(\cos 0 = \Large{-1 \over 2}\)
that's how you do what i said "take the arc cos of both sides"
\[\cos 0 = 1 \ne {-1 \over 2} \]
who said it was -1/2 o.O
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The question.
i meant who said cos 0 = -1/2?
Your solution says \(x = 0\).
lol look at the start of my solution
I see.
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You still can't evaluate \(\cos ^{-1}({-1 \over 2})\) without using a calculator/unit circle.
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