Question

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

Answer 1

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.