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Question: If \(2{\sin ^2}\theta - 5\sin \theta + 2 > 0,\,\theta \in \left( {0,2\pi } \right)\), then \(\theta ...

If 2sin⁡2θ−5sin⁡θ+2>0, θ∈(0,2π)2{\sin ^2}\theta - 5\sin \theta + 2 > 0,\,\theta \in \left( {0,2\pi } \right), then θ∈\theta \in
A. (5π6,2π)\left( {\dfrac{{5\pi }}{6},2\pi } \right)
B. (0,π6)∪(5π6,2π)\left( {0,\dfrac{\pi }{6}} \right) \cup \left( {\dfrac{{5\pi }}{6},2\pi } \right)
C. (0,π6)\left( {0,\dfrac{\pi }{6}} \right)
D. (π80,π6)\left( {\dfrac{\pi }{{80}},\dfrac{\pi }{6}} \right)

Explanation

Solution

First, substitute a variable at the place of sin⁡x\sin x. Now factor the equation and find the roots of the equation. Now, substitute back sin⁡x\sin x in place of the variable. Cancel out one factor as sin⁡x\sin x cannot be greater than 1. Now find the interval where sin⁡x\sin x is less than 12\dfrac{1}{2}. Also include the 3rd3^{rd} and 4th4^{th} interval, as the value of sin⁡x\sin x is negative in that interval. The interval derived is the final answer.

Complete step-by-step solution:
Given: - 2sin⁡2θ−5sin⁡θ+2>02{\sin ^2}\theta - 5\sin \theta + 2 > 0
Let u=sin⁡θu = \sin \theta . So,
2u2−5u+2>02{u^2} - 5u + 2 > 0
Now, factor the equation on the left side,
⇒\Rightarrow 2u2−4u−u+2>02{u^2} - 4u - u + 2 > 0
Factor out 2u from the first group and -1 from the second group,
⇒\Rightarrow 2u(u−2)−1(u−2)>02u\left( {u - 2} \right) - 1\left( {u - 2} \right) > 0
Now factor out (u-2) from both groups,
⇒\Rightarrow (2u−1)(u−2)>0\left( {2u - 1} \right)\left( {u - 2} \right) > 0
As we know that, if (x−a)(x−b)>0\left( {x - a} \right)\left( {x - b} \right) > 0 and a<ba < b. Then, x<ax < a or x>bx > b. So,
⇒\Rightarrow 2u−1<02u - 1 < 0 or u−2>0u - 2 > 0
Move the constant on the right side,
⇒\Rightarrow 2u<12u < 1 or u>2 u > 2
Divide 2u<12u < 1 by 2,
⇒\Rightarrow u<12u < \dfrac{1}{2} or u>2 u > 2
Now, substitute back sin⁡x\sin x in place of u.
⇒\Rightarrow sin⁡x<12\sin x < \dfrac{1}{2} or sin⁡x>2 \sin x > 2
Since the value of sin⁡x\sin x oscillates between -1 and 1. So, sin⁡x\sin x cannot be greater than 1. Thus, discard sin⁡x>2\sin x > 2.
Then,
sin⁡x<12\sin x < \dfrac{1}{2}
Since, sin⁡x=sin⁡π6\sin x = \sin \dfrac{\pi }{6} and sin⁡x=sin⁡5π6\sin x = \sin \dfrac{{5\pi }}{6}. Then,
x<π6x < \dfrac{\pi }{6} or x>5π6x > \dfrac{{5\pi }}{6}.
Since the value of sin⁡x\sin x is negative in the 3rd and 4th quadrant. So,
x∈(0,π6)x \in \left( {0,\dfrac{\pi }{6}} \right) or x∈(5π6,2π)x \in \left( {\dfrac{{5\pi }}{6},2\pi } \right)
Thus, x∈(0,π6)∪(5π6,2π)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 (x−2)(x−5)<0\left( {x - 2} \right)\left( {x - 5} \right) < 0, then 2<x<52 < x < 5.
If (x−2)(x−6)>0\left( {x - 2} \right)\left( {x - 6} \right) > 0, then x<2x < 2 and x>6x > 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.