find all possible solutions 
The set of all $x$ in the interval $[0,\pi]$ for which $2\sin^2x-3\sin x+1 \geq 0$, is _________________.

I have tried by factoring it first and then comparing it with the inequality.
My final step was $(\sin x-1)(2\sin x-1) \geq 0$.
 A: Hint:  Note that $\sin x -1 \lt 0$ except at one point in the interval $[0,\pi]$.  Where $\sin x -1 = 0$ equals zero, the required inequality is satisfied.  Otherwise, you need $2 \sin x -1 \le 0$  Can you solve that?
A: let $y = \sin t.$ you have $$2y^2 -3y + 1 = (2y-1)(y-1) \ge 0 $$ that is $$ -1 \le y \le \frac12, \, y = 1.$$  the range for $t$ is $$0 \le t \le \pi/6, \quad \pi/2, \quad 5\pi/6 \le t \le 2\pi. $$
A: As given, you can factorize the expression, as follows $$2\sin^2x-3\sin x+1\geq 0 \implies (2\sin x-1)(\sin x-1)\geq 0$$ $$\implies \left(\sin x-\frac{1}{2}\right)(\sin x-1)\geq 0$$ On solving the above inequality for $\sin x$, we get  $$\sin x\leq \frac{1}{2}\space \text{or} \space \sin x\geq 1  $$ Now, by solving the first part for the given interval $[0, \pi]$, we get two sets of the values of $x$ as follows $$0\leq x\leq \frac{\pi}{6}\space \text{&} \space \frac{5\pi}{6}\leq x\leq \pi$$ Similarly, by solving the second part for the given interval $[0, \pi]$, we get the value of $x$ as follows $$x=\frac{\pi}{2}$$ Hence, by writing the complete solution of the given inequality, we have $$x\in \left[0, \frac{\pi}{6}\right] \cup \{\frac{\pi}{2}\} \cup \left[\frac{5\pi}{6}, \pi\right] $$
A: This trinomial is continuous on your interval so set it to zero, find the roots, from which you can make a table, and read off the values of $x$ for which it is $+$ or $-$.
Hint: the roots are $\frac{\pi }{2}$ and $\frac{\pi }{6}$.
