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Shaydenparis
Group Title
solve the following equation on interval [0,2pi) 1cosx=sinx
 one year ago
 one year ago
Shaydenparis Group Title
solve the following equation on interval [0,2pi) 1cosx=sinx
 one year ago
 one year ago

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whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
Let's write c instead of \(\cos x\) and s instead of \(\sin x\). This is just to reduce the amount of writing involved. \[1c = s\]Square both sides \[(1c)(1c) = 12c+c^2 = s^2\]Remember that \[\sin^2x+\cos^2x = 1\]so we can rewrite the right hand side as \[12c+c^2=1c^2\]If we solve that for \(c\) \[2c+2c^2 = 0\]\[c+c^2=0\]\[c^2=c\]\[c=1\]And now we undo our substitution to reveal\[\cos x = 1\]Where does \(\cos x=1\) in the interval \([0,2\pi)\)?
 one year ago

Shaydenparis Group TitleBest ResponseYou've already chosen the best response.0
thank you so much
 one year ago
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