Question

The equation $4{{\sin }^{2}}x+4\sin x+{{a}^{2}}-3=0$ posses 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: For this problem we need to calculate the range of the variable $a$ if the given equation has a solution. We can observe that the given equation is a quadratic equation in terms of $\sin x$. We will calculate the solution of the given quadratic equation by using the quadratic formula which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. So, we will first compare the given equation with $a{{x}^{2}}+bx+c$ and calculate the roots with the formula. In the problem they have mentioned that the given equation has a solution, so we will assume the range of the solution and simplify the equation to get the required result.

Complete step-by-step solution:
Given the equation, $4{{\sin }^{2}}x+4\sin x+{{a}^{2}}-3=0$.
We can observe that the above equation is a quadratic equation in terms of $\sin x$. Now comparing the above equation with $a{{x}^{2}}+bx+c$, then we will get
$a=4$, $b=4$, $c={{a}^{2}}-3$.
Now the solution of the given equation will be
$\begin{align}
  & \Rightarrow \sin x=\dfrac{-4\pm \sqrt{{{4}^{2}}-4\left( 4 \right)\left( {{a}^{2}}-3 \right)}}{2\left( 4 \right)} \\
 & \Rightarrow \sin x=\dfrac{-4\pm \sqrt{16-16{{a}^{2}}+48}}{8} \\
 & \Rightarrow \sin x=\dfrac{-4\pm \sqrt{64-16{{a}^{2}}}}{8} \\
\end{align}$
Taking $16$ as common from the value $\sqrt{64-16{{a}^{2}}}$, then we will get
$\Rightarrow \sin x=\dfrac{-4\pm 4\sqrt{4-{{a}^{2}}}}{8}$
To have the value of above solution we need to have $4-{{a}^{2}}\ge 0$. Considering this condition and simplifying it, then we will get
$\begin{align}
  & \Rightarrow 4-{{a}^{2}}\ge 0 \\
 & \Rightarrow {{2}^{2}}-{{a}^{2}}\ge 0 \\
\end{align}$
Applying the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the above equation, then we will get
$\begin{align}
  & \Rightarrow \left( 2+a \right)\left( 2-a \right)\ge 0 \\
 & \Rightarrow 2+a\ge 0\text{ and }2-a\ge 0 \\
 & \Rightarrow a\ge -2\text{ and }a\le 2 \\
\end{align}$
From the above conditions we can write the range of the variable $a$ as $\left[ -2,2 \right]$.

Note: We can also plot the graph of the given equation by taking random values of $a$ and checking whether the given equation has a solution or not. After getting the list of values of $a$ for which we have a solution for the given equation we can decide the range of the variable.