# Find all of the solutions of $2\sin(t)−1−\sin^2(t)=0$ in the interval $[0,2\pi]$

Find all of the solutions of $2\sin(t)−1−\sin^2(t)=0$ in the interval $[0,2\pi]$.

Here is my work so far:

$2\sin(t)−1−\sin^2(t)=0$

$2\sin(t)−\sin^2(t)-1=0$

Using the identity: $\sin^2\theta -1=-\cos^2\theta$

$2\sin(t)+\cos^2(t)$

Where do I go from here? Any advice or hints would be appreciated.

• How did you get $2\sin(t)\cos^2(t)$ using that identity? And what is it equal to? 0? – Thad Janisse Oct 12 '15 at 2:58
• Any chance you mean "\sin^2(t)" where you have "sin2(t)"? and "2 \sin(t) - \cos^2(t) = 0" in your last non-text line? – Eric Towers Oct 12 '15 at 3:00
• @ThadJanisse Just edited the problem, I had forgotten the exponent.. Sorry about that. – McB Oct 12 '15 at 3:00
• That is exactly what I meant, @EricTowers! Sorry about that. – McB Oct 12 '15 at 3:01

This is a quadratic in $\sin(t)$. It might be easier to see as:

Find all of the solutions to $2u-1-u^2 = 0$ for $u = \sin(\theta)$ and $\theta$ in $[0,2\pi]$.

Write $x = \sin t$, then we have

$$2x - 1 - x^2 =0$$

which is equivalent to $x^2-2x+1 = 0$ which is equivalent to $(x-1)^2 = 0$ which holds when $x = 1$. So, $\sin t = 1$ iff $t = \frac{\pi}{2}$ on the interval $[0, 2 \pi]$

$$\begin{gathered} Eq \Leftrightarrow \left\{ \begin{gathered} {\sin ^2}t - 2\sin t + 1 = 0 \hfill \\ 0 \leqslant t \leqslant 2\pi \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} {\left( {\sin t - 1} \right)^2} = 0 \hfill \\ 0 \leqslant t \leqslant 2\pi \hfill \\ \end{gathered} \right. \hfill \\ \Leftrightarrow \left\{ \begin{gathered} \sin t = 1 \hfill \\ 0 \leqslant t \leqslant 2\pi \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} t = \frac{\pi } {2} + k2\pi \hfill \\ 0 \leqslant t \leqslant 2\pi \hfill \\ \end{gathered} \right. \Leftrightarrow t = \frac{\pi } {2}. \hfill \\ \end{gathered}$$

Try the substitution $t = \arcsin(x)$. This should make things easier to solve. Once you find an $x$ (or $x$'s) simply evaluate $\arcsin(x)$.