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# The equation $4{\sin ^2}x + 4\sin x + {a^2} - 3 = 0$ possesses a solution if ‘a’ belongs to the intervalA. $\left( { - 1,3} \right)$B. $\left( { - 3,1} \right)$C. $\left( { - 2,2} \right)$D. $R - \left( { - 2,2} \right)$

Last updated date: 17th Jun 2024
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Answer
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Hint: We solve the equation for the value of ‘a’ by converting the part of the given equation similar to the formula${(a + b)^2} = {a^2} + {b^2} + 2ab$. Since we know a solution of an equation exists implies its factors are equal to zero, so we equate the factors to zero and find the value of ‘a’.

Complete step-by-step solution:
We are given the equation$4{\sin ^2}x + 4\sin x + {a^2} - 3 = 0$
We add and subtract the same term from LHS of the equation such that it helps us to pair up terms in a way like the identity${(a + b)^2} = {a^2} + {b^2} + 2ab$.
Add and subtract 1 from LHS of the equation.
$\Rightarrow 4{\sin ^2}x + 4\sin x + {a^2} - 3 + 1 - 1 = 0$....................… (1)
We can write;${1^2} = 1$,$4{\sin ^2}x = {(2\sin x)^2}$and$4\sin x = 2 \times 2\sin x \times 1$
Substitute these values in equation (1)
$\Rightarrow {\left( {2\sin x} \right)^2} + {(1)^2} + 2 \times \left( {2\sin x \times 1} \right) + {a^2} - 3 - 1 = 0$...............… (2)
Use the identity ${(a + b)^2} = {a^2} + {b^2} + 2ab$to write${\left( {2\sin x} \right)^2} + {(1)^2} + 2 \times \left( {2\sin x \times 1} \right) = {\left( {2\sin x + 1} \right)^2}$
Substitute the value of ${\left( {2\sin x} \right)^2} + {(1)^2} + 2 \times \left( {2\sin x \times 1} \right) = {\left( {2\sin x + 1} \right)^2}$in equation (2)
$\Rightarrow {\left( {2\sin x + 1} \right)^2} + {a^2} - 4 = 0$
Shift all constant values to RHS of the equation
$\Rightarrow {\left( {2\sin x + 1} \right)^2} = 4 - {a^2}$...................… (3)
For equation (1) to have a solution RHS should be equal to zero
$\Rightarrow 4 - {a^2} = 0$
Shift constant value to one side of the equation
$\Rightarrow 4 = {a^2}$
Take square root on both sides of the equation
$\Rightarrow \sqrt 4 = \sqrt {{a^2}}$
$\Rightarrow \sqrt {{2^2}} = \sqrt {{a^2}}$
Cancel square root by square power on both sides of the equation
$\Rightarrow a = \pm 2$
$\Rightarrow a = - 2,2$...................… (4)
$\therefore a \in ( - 2,2)$

$\therefore$Option C is the correct answer.

Note: Alternate method:
We can also find the value using a discriminant method.
Since Discriminant of a quadratic equation $a{x^2} + bx + c = 0$is given by $D = {b^2} - 4ac$
On comparing equation$4{\sin ^2}x + 4\sin x + {a^2} - 3 = 0$ with general quadratic equation;
$a = 4;b = 4;c = {a^2} - 3$
Discriminant of equation$4{\sin ^2}x + 4\sin x + {a^2} - 3 = 0$ will be
$\Rightarrow D = {(4)^2} - 4(4)({a^2} - 3)$
$\Rightarrow D = 16 - 16{a^2} + 48$
Add constant terms
$\Rightarrow D = - 16{a^2} + 64$.................… (1)
We know if $D \geqslant 0$ then the equation has two real solutions.
We substitute the value of D from equation (1)
$\Rightarrow - 16{a^2} + 64 \geqslant 0$
Take -16 common from both terms
$\Rightarrow - 16({a^2} - 4) \geqslant 0$
$\Rightarrow ({a^2} - 4) \geqslant 0$
Since we know the identity ${x^2} - {y^2} = (x - y)(x + y)$
$\Rightarrow (a - 2)(a + 2) \geqslant 0$
Either $a \geqslant 2$or $a \leqslant - 2$
$\therefore a \in ( - 2,2)$
$\therefore$Option C is correct.