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1/3 = 1/2 Cos^2 (theta), solve for theta
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can someone solve this, showing ALL steps, i am very rusty on my trig functions/algebra
I will post a response soon. If my browser lets me
thks!
\[\frac{1}{3} = \frac{\cos^2(\theta)}{2}\]
\[3\cos^2(\theta) = 2\]
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\[\cos^2(\theta) = \frac{2}{3}\]
\[\cos(\theta)^2 = \frac{2}{3}\]
\[\cos(\theta) = \pm \sqrt{\frac{2}{3}}\]
\[\theta = \cos^{-1}(\pm \sqrt{\frac{2}{3}})\]
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yes! you are amazing!
What values did you get for theta?
Thanks for the testimonial by the way :)
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