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Question: Find the number of solutions of the equation \(2{({\sin ^{ - 1}}x)^2} - {\sin ^{ - 1}}x - 6 = 0\). ...

Find the number of solutions of the equation 2(sin⁡−1x)2−sin⁡−1x−6=02{({\sin ^{ - 1}}x)^2} - {\sin ^{ - 1}}x - 6 = 0.
A) 22
B) 11
C) 00
D) 33

Explanation

Solution

The given equation is a second-degree equation in sin⁡−1x{\sin ^{ - 1}}x. To find what all values xx can take (and thus the number of solutions), first we need to check what all values sin⁡−1x{\sin ^{ - 1}}x can take. This can be done by solving the quadratic equation by taking sin⁡−1x=y{\sin ^{ - 1}}x = y(say). Then we can eliminate any values which does not belong to the range of sin⁡−1x{\sin ^{ - 1}}x.

Formula used:
A second-degree equation of the form ax2+bx+c=0a{x^2} + bx + c = 0 can be solved by
x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

Complete step by step answer:
The given quadratic equation is 2(sin⁡−1x)2−sin⁡−1x−6=02{({\sin ^{ - 1}}x)^2} - {\sin ^{ - 1}}x - 6 = 0
We are asked to find the number of solutions.
Here the variable present is xx. So, we need to find the number of values xx can take.
Let, sin⁡−1x=y{\sin ^{ - 1}}x = y
Then the equation becomes,
⇒2y2−y−6=0\Rightarrow 2{y^2} - y - 6 = 0, which is a quadratic equation in yy.
A second-degree equation of the form ax2+bx+c=0a{x^2} + bx + c = 0 can be solved by
⇒x=−b±b2−4ac2a\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Here the variable is yy.
Also a=2,b=−1,c=−6a = 2,b = - 1,c = - 6
⇒y=−(−1)±(−1)2−4×2×−62×2\Rightarrow y = \dfrac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4 \times 2 \times - 6} }}{{2 \times 2}}
Simplifying we get,
⇒y=1±1+484=1±494\Rightarrow y = \dfrac{{1 \pm \sqrt {1 + 48} }}{4} = \dfrac{{1 \pm \sqrt {49} }}{4}
⇒y=1±74\Rightarrow y = \dfrac{{1 \pm 7}}{4}
This gives y=84=2y = \dfrac{8}{4} = 2 or y=−64=−32y = \dfrac{{ - 6}}{4} = \dfrac{{ - 3}}{2}
Substituting for yy we get,
⇒sin⁡−1x=2 (or) sin⁡−1x=−32\Rightarrow {\sin ^{ - 1}}x = 2{\text{ (or) }}{\sin ^{ - 1}}x = \dfrac{{ - 3}}{2}
But sin⁡−1x∈(−π2,π2){\sin ^{ - 1}}x \in ( - \dfrac{\pi }{2},\dfrac{\pi }{2}), which is the open interval with limits −π2,π2 - \dfrac{\pi }{2},\dfrac{\pi }{2}.
Here π\pi is in radians, which can be approximated to 3.143.14.
This gives
⇒π2=3.142=1.57\Rightarrow \dfrac{\pi }{2} = \dfrac{{3.14}}{2} = 1.57
So sin⁡−1x{\sin ^{ - 1}}x cannot take the value 22, since the value must be less than 1.571.57.
Therefore sin⁡−1x=−32⇒x=sin⁡(−32){\sin ^{ - 1}}x = \dfrac{{ - 3}}{2} \Rightarrow x = \sin (\dfrac{{ - 3}}{2}), which is the only solution.
This gives the number of solutions is 11.

∴\therefore The correct answer is option B.

Note:
The alternative method to solve the quadratic equation without this formula.
2y2−y−6=02{y^2} - y - 6 = 0
By simple rearrangement we get,
⇒2y2−4y+3y−6=0\Rightarrow 2{y^2} - 4y + 3y - 6 = 0
Taking common factors from first two terms and last two terms,
⇒2y(y−2)+3(y−2)=0\Rightarrow 2y(y - 2) + 3(y - 2) = 0
⇒(y−2)(2y+3)=0\Rightarrow (y - 2)(2y + 3) = 0
The product of two terms is zero implies either one is zero.
⇒(y−2)=0 (or) (2y+3)=0\Rightarrow (y - 2) = 0{\text{ (or) }}(2y + 3) = 0
⇒y=2(or)=−32\Rightarrow y = 2\left( {{\text{or}}} \right) = \dfrac{{ - 3}}{2}