Question

If $ 4{{\sin }^{2}}\left( x \right)-8\sin \left( x \right)+3\le 0 $ , then the solution set for x $ 0\le x\le 2\pi $ is
(A) $ \left[ 0,\dfrac{\pi }{6} \right] $
(B) $ \left[ 0,\dfrac{5\pi }{6} \right] $
(C) $ \left[ \dfrac{5\pi }{6},2\pi \right] $
(D) $ \left[ \dfrac{\pi }{6},\dfrac{5\pi }{6} \right] $

Answer 1

Hint: We will solve this question by first solving the quadratic equation by either method, that is, by quadratic formula or by splitting the middle term method. Then we will check the domain and range of Sin[x] so as to find the value of x. We will use $ -1\le \sin \left( x \right)\le 1 $ to find the solution set for x. The concept that if $ \left( a \right)\left( b \right)\le 0 $ , then either $ a\le 0 $ or $ b\le 0 $ but not both are less than zero at the same time.

Complete step-by-step answer:
We are given the equation, $ 4{{\sin }^{2}}\left( x \right)-8\sin \left( x \right)+3\le 0 $ , which we have to solve first.
It can be solved by two methods, we solve this quadratic equation by splitting the middle term method, that is,
 $ 4{{\sin }^{2}}\left( x \right)-8\sin \left( x \right)+3\le 0 $
Now we factorize the middle term, that is coefficient of x (b), in two terms such that their sum is equal to middle term, that is coefficient of x and the product of these 2 numbers is equal to product of first and last term, that is, coefficient of $ {{x}^{2}} $ (a) and constant term (c), where $ a{{x}^{2}}+bx+c=0 $ is general equation.
Now, comparing the given equation $ 4{{\sin }^{2}}\left( x \right)-8\sin \left( x \right)+3\le 0 $ with general equation $ a{{x}^{2}}+bx+c=0 $ , we get
 $ x=\sin \left( x \right) $
a = 4
b = -8
c = 3
Now, applying splitting the middle term method,
 $ 4{{\sin }^{2}}\left( x \right)-6\sin \left( x \right)-2\sin \left( x \right)+3\le 0 $
 $ 2\sin \left( x \right)\left[ 2\sin \left( x \right)-3 \right]-1\left[ 2\sin \left( x \right)-3 \right]\le 0 $
 $ \left[ 2\sin \left( x \right)-3 \right]\left[ 2\sin \left( x \right)-1 \right]\le 0 $
We know that,
  $ -1\le \sin \left( x \right)\le 1 $
This implies that $ 2\sin \left( x \right)-3\le 0 $ , which means that $ 2\sin \left( x \right)-1\ge 0 $ as if $ \left( a \right)\left( b \right)\le 0 $ , then either $ a\le 0 $ or $ b\le 0 $ but not both are less than zero at the same time.
 $ 2\sin \left( x \right)-1\ge 0 $ , which implies that $ \sin \left( x \right)\ge \dfrac{1}{2} $
And $ -1\le \sin \left( x \right)\le 1 $ , also $ -1\le \dfrac{1}{2} $
Which implies that $ \dfrac{1}{2}\le \sin \left( x \right)\le 1 $
Therefore, the solution set for x should be $ \left[ \dfrac{\pi }{6},\dfrac{5\pi }{6} \right] $
So, the correct answer is “Option D”.

Note: This question involves understanding of trigonometric values, domain and range of trigonometric functions. This question also involves the concepts of algebra which are required to solve the quadratic equation. The quadratic equation can be solved by either method by quadratic formula or by splitting the middle term method. Take care while doing the calculations.