Solve for $\displaystyle 0\leq x \leq 2\pi$ well i am bad at the math tool, but its 0<=x<=2pi Cos2x = 1 - sinx What would be the best way to set this up?
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Originally Posted by adam182 Solve for $\displaystyle 0\leq x \leq 2\pi$ well i am bad at the math tool, but its 0<=x<=2pi Cos2x = 1 - sinx What would be the best way to set this up? Hi Adam. Use the identity that $\displaystyle \cos(2x) = 1 - 2\sin^2x$. Get: $\displaystyle 1-2\sin^2x = 1 - \sin x$. So $\displaystyle 2 \sin^2 x - \sin x = 0$. Factor: $\displaystyle \sin x \cdot (2 \sin x - 1) = 0$. Can you finish it from here?
Yes, I got to x=30 which I converted to radians...so $\displaystyle \pi/6$
Yes $\displaystyle \pi/6$ is a solution. There are several answers in that interval given $\displaystyle (0 \le x \le 2\pi)$. Note that $\displaystyle 5\pi/6$ is also an answer, as are $\displaystyle 0, \pi$, and $\displaystyle 2\pi$. Do you see why?
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