I'm stuck on this problem...
Find all solutions in the interval [0,2π]: 1 - sinx = 2 cos^2x
any help would be appreciated.
Hello adam456006Use the identity $\displaystyle \cos^2x = 1-\sin^2x$ and then re-arrange as a quadratic in $\displaystyle \sin x$:
$\displaystyle 1 - \sin x = 2(1-\sin^2x)$
$\displaystyle \Rightarrow 2\sin^2 x -\sin x -1 = 0$
$\displaystyle \Rightarrow (\sin x - 1)(2\sin x +1) = 0$
$\displaystyle \Rightarrow \sin x = 1$ or $\displaystyle \sin x = -\tfrac{1}{2}$
I'll leave these to you.
Grandad