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find the value of θ in the equation cos2θ+sin²θ=cos³θ+3θ-6
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i got as far as substituting the double angle formula in there and simplifying to get cos^2x=cos^3x+3x-6 cos^2x-cos^3x=3x-6
yeah that is how far i got too
you could try solving \(u^3-u^2+3u-6=0\) and and then take the inverse cosine if you can find such a solution
okay well its a multiple choice: so x could equal 6 cos2x x or 2
then cheat http://www.wolframalpha.com/input/?i=cos%282x%29%2Bsin^2%28x%29%3Dcos^3%28x%29%2B3x-6
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2 it is then lol thanks!
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