Find all solutions of $2\sin t-1-\sin^2 t = 0$ in $[0,2\pi]$.
Attempt: \begin{align} 1-2\sin t& =-\sin^2 t\\ \cos 2t& =-\sin^2 t\\ \cos^2t-\sin^2t & =-\sin^2t\\ \cos^2t & = 0. \end{align} Am I correct up to this point? If I have done everything right, I do not know where to go from here. I know how to solve $\cos t=0$ but $\cos^2$ is confusing me. Any help would be appreciated, thanks!
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$$ \begin{array}{llll} 2sin(t)-1-sin^2(t)=0\implies 0=\stackrel{\textit{notice, is just }ax^2+bx+c}{sin^2(t)-2sin(t)+1} \\ \quad \\ 0=[sin(t)-1][sin(t)-1]\implies 1=sin(t) \\ \quad \\ sin^{-1}(1)=sin^{-1}[sin(t)]\implies sin^{-1}(1)=t\implies {\color{blue}{ \cfrac{\pi }{2}=t }} \end{array} $$
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It’s quite simple to solve. Although not necessary, simply substitute $\sin (t)$ with $t$ or any other variable, which leads to a normal quadratic equation.
$2\sin (t)-1-\sin^2 (t)$ = $0$
$2t-1-t^2$ = $0$
$-t^2+2t-1$ = $0$ $\Rightarrow$ $t^2-2t+1$ = $0$ $\Rightarrow$ $(t-1)^2$ = $0$ $\Rightarrow$ $t-1$ = $0$ $\Rightarrow$ $t$ = $1$
Now, simply plug in $\sin (t)$ = $t$.
$\sin (t)$ = $1$ $\Rightarrow$ $t$ = $\sin^-1 (1)$ = $\frac{\pi}{2}$
However, the question asks for a complete set of solutions, rather than those for $0\leq t \leq 2\pi$, so our complete answer is $t$ = $\frac{\pi}{2} + 2\pi n$ for all integer values of $n$.
(You made a mistake when attempting to simplify $1-2\sin (t)$ = $-\sin^2 (t)$. $\cos (2t)$ = $1-2\sin^2 (t)$, NOT $1-2\sin (t)$.)
Substitute $$\sin(t)=z$$ then we get $$2z-1-z^2=0$$ which is easy to solve.