sin^2 theta + cos theta = 2 (Hint: Use the Pythagorean identity sin^2 theta + cost^2 theta = 1 to replace sin^2 theta in the given equation.)
\[ \sin^2 \theta + \cos^2 \theta = 1 \\ \sin^2 \theta = 1-\cos^2 \theta\]
\[1-\cos^2\Theta = 2\] So then it would be this^?
you start with \[ \sin^2 \theta + \cos \theta = 2 \] replace the sin^2 with 1 - cos^2 in that equation. what do you get ?
\[1-\cos^2\Theta+cosTheta=2\]
you get \[ 1 - \cos^2 \theta + \cos \theta = 2\] which simplifies to \[ - \cos^2 \theta + \cos \theta -1 =0 \\ \cos^2 \theta - \cos \theta +1 =0 \] if you let x= cosθ you have \[ x^2 -x +1 = 0\] you will find x= (1± sqr(3) i) /2 this is an imaginary number. That means there is no real solution. So you are doing a complicated problem or you have a typo in the problem
Yeah, maybe there's a typo x)
Thanks for your help!
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