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Medals for you! can i have some help from the people at open study? solve each equation 1.2cos^2(theta)-3cos(theta) +1=0 , 0= (theta) < 2pi
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@SithsAndGiggles would you be able to help me?
It's a quadratic in disguise. Try writing \(x=\cos\theta\) to make it easier to see: \[2x^2-3x+1=0\]
wow thats so much clearer...good!
x=1?
Not quite, you can factor the expression: \[2x^2-3x+1=(2x-1)(x-1)=0\] which gives \(x=1\) *and* \(x=\dfrac{1}{2}\). Remember that \(x=\cos\theta\), so in fact you have \[\cos\theta=1~~\text{and}~~\cos\theta=\frac{1}{2}\]
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right, which makes a lot more sense..this is great, honestly thanks for helping me with such a simple one! thanks XD
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