Question

If $2{\sin ^2}\theta - 5\sin \theta + 2 > 0,\,\theta \in \left( {0,2\pi } \right)$, then $\theta \in $
A. $\left( {\dfrac{{5\pi }}{6},2\pi } \right)$
B. $\left( {0,\dfrac{\pi }{6}} \right) \cup \left( {\dfrac{{5\pi }}{6},2\pi } \right)$
C. $\left( {0,\dfrac{\pi }{6}} \right)$
D. $\left( {\dfrac{\pi }{{80}},\dfrac{\pi }{6}} \right)$

Answer 1

Hint: First, substitute a variable at the place of $\sin x$. Now factor the equation and find the roots of the equation. Now, substitute back $\sin x$ in place of the variable. Cancel out one factor as $\sin x$ cannot be greater than 1. Now find the interval where $\sin x$ is less than $\dfrac{1}{2}$. Also include the $3^{rd}$ and $4^{th}$ interval, as the value of $\sin x$ is negative in that interval. The interval derived is the final answer.

Complete step-by-step solution:
Given: - $2{\sin ^2}\theta - 5\sin \theta + 2 > 0$
Let $u = \sin \theta $. So,
$2{u^2} - 5u + 2 > 0$
Now, factor the equation on the left side,
$\Rightarrow$$2{u^2} - 4u - u + 2 > 0$
Factor out 2u from the first group and -1 from the second group,
$\Rightarrow$$2u\left( {u - 2} \right) - 1\left( {u - 2} \right) > 0$
Now factor out (u-2) from both groups,
$\Rightarrow$$\left( {2u - 1} \right)\left( {u - 2} \right) > 0$
As we know that, if $\left( {x - a} \right)\left( {x - b} \right) > 0$ and $a < b$. Then, $x < a$ or $x > b$. So,
$\Rightarrow$$2u - 1 < 0 $ or $u - 2 > 0$
Move the constant on the right side,
$\Rightarrow$$2u < 1 $ or $ u > 2$
Divide $2u < 1$ by 2,
$\Rightarrow$$u < \dfrac{1}{2} $ or $ u > 2$
Now, substitute back $\sin x$ in place of u.
$\Rightarrow$$\sin x < \dfrac{1}{2} $ or $ \sin x > 2$
Since the value of $\sin x$ oscillates between -1 and 1. So, $\sin x$ cannot be greater than 1. Thus, discard $\sin x > 2$.
Then,
$\sin x < \dfrac{1}{2}$
Since, $\sin x = \sin \dfrac{\pi }{6}$ and $\sin x = \sin \dfrac{{5\pi }}{6}$. Then,
$x < \dfrac{\pi }{6}$ or $x > \dfrac{{5\pi }}{6}$.
Since the value of $\sin x$ is negative in the 3rd and 4th quadrant. So,
$x \in \left( {0,\dfrac{\pi }{6}} \right)$ or $x \in \left( {\dfrac{{5\pi }}{6},2\pi } \right)$
Thus, $x \in \left( {0,\dfrac{\pi }{6}} \right) \cup \left( {\dfrac{{5\pi }}{6},2\pi } \right)$

Hence, option (B) is the correct answer.

Note: The students are likely to make mistakes in questions when finding the roots for > sign.
For e.g.,
If $\left( {x - 2} \right)\left( {x - 5} \right) < 0$, then $2 < x < 5$.
If $\left( {x - 2} \right)\left( {x - 6} \right) > 0$, then $x < 2$ and $x > 6$.
Also, keep in mind the number of factors will never exceed the highest power of the polynomial. Also, students can always check if the factors are correct or not by substituting the values in the expression, the expression turns out to be zero when we substitute the correct values.