Question

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

Answer 1

To solve this, first find the roots for $\sin \theta$ consider it as the quadratic equation and the find the roots
Secondly, use the limit to find the value $\theta \in$.
Here we go.

Complete step-by-step answer:
It is given that, $2{\sin ^2}\theta - 5\sin \theta + 2 > 0,\theta \in (0,2\pi )$
Consider,
$2{\sin ^2}\theta - 5\sin \theta + 2 = 0$
Take $\sin \theta = x$
Then, $2{x^2} - 5x + 2 = 0$
Using the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ for $a{x^2} + bx + c$
Here, $b = - 5$,$a = 2$,$c = 2$
Substitute the value in the formula, we get
$x = \dfrac{{ - ( - 5) \pm \sqrt {{5^2} - 4(2)(2)} }}{{2(2)}}$
Minus of minus is plus
$x = \dfrac{{(5) \pm \sqrt {{5^2} - 4(2)(2)} }}{{2(2)}}$
On squaring the inner terms and multiply that we get,
$x = \dfrac{{5 \pm \sqrt {25 - 16} }}{4}$
Let us subtract the square root terms, we get
$x = \dfrac{{5 \pm \sqrt 9 }}{4}$
Taking the square root of $\sqrt 9$ is$3$
$x = \dfrac{{5 \pm 3}}{4}$
Separate the terms
$x = \dfrac{{5 + 3}}{4},\dfrac{{5 - 3}}{4}$
Further simplify,
$x = \dfrac{8}{4},\dfrac{2}{4}$
On dividing the terms we get,
$x = 2,\dfrac{1}{2}$
Substitute $x = \sin \theta$
We can write it as, $\sin \theta = 2,\dfrac{1}{2}$
For four given we can write, equal to is substitute by greater symbol
$\sin \theta > 2,\dfrac{1}{2}$
Also we can write it as,
$(\sin \theta - 2)(\sin \theta - \dfrac{1}{2}) > 0$
Here, in trigonometry, $\sin \theta$ values lie in between $[ - 1,1]$
$(\sin \theta - 2)$ lie in between$[ - 1 - 2,1 - 2]$
That is, $(\sin \theta - 2)$ lie in between $[ - 3, - 1]$
We can write this as $- 3 \leqslant \sin \theta - 2 \leqslant - 1$
Next we can take $- 1 \leqslant \sin \theta \leqslant 1$
$- 1 - \dfrac{1}{2} \leqslant \sin \theta - \dfrac{1}{2} \leqslant 1 - \dfrac{1}{2}$
$- \dfrac{3}{2} \leqslant \sin \theta - \dfrac{1}{2} \leqslant \dfrac{1}{2}$
Then,
The value $(\sin \theta - 2)$ will never be zero
So, $(\sin \theta - \dfrac{1}{2})$ is less than $0$
That is, $(\sin \theta - \dfrac{1}{2}) < 0$
$\therefore$ We can write it as, $\sin \theta < \dfrac{1}{2}$
Take the $\sin$ left hand side to right hand side, that is, we can write it as inverse sign
$\theta < {\sin ^{ - 1}}\dfrac{1}{2}$
${\sin ^{ - 1}}\dfrac{1}{2} = \dfrac{\pi }{6}$
$\theta < \dfrac{\pi }{6}$
If we draw a straight line to $x = 0.5$ then the graph will cut the points $30^\circ$ and ${150^\circ }$,
Given, $\theta \in (0,2\pi )$
Here we take $\pi = {180^ \circ }$
We find $\theta < \dfrac{\pi }{6}$ then,
\left( {{0^\circ },{{30}^\circ }} \right)$$$$({150^\circ },{360^\circ })
By radians we can write it as,
$\left( {0.\dfrac{\pi }{6}} \right) \cup \left( {\dfrac{{5\pi }}{6},2\pi } \right)$

So, the correct answer is “Option C”.

Note: Here, the sum we solve is the long method, we won’t solve like this, but here we want to understand with basics, so that we solve it with explanations each and every step.
In trigonometric function $\sin \theta ,\cos \theta$ value lies between $[ - 1,1]$ it is an important note, except that others all have different limits.
These types of sum first find the limits of a given trigonometric function; it is useful to negotiate unwanted values.