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Question: The equation \(4{{\sin }^{2}}x+4\sin x+{{a}^{2}}-3=0\) posses a solution if ‘\(a\)’ belongs to the i...

The equation 4sin⁡2x+4sin⁡x+a2−3=04{{\sin }^{2}}x+4\sin x+{{a}^{2}}-3=0 posses a solution if ‘aa’ belongs to the interval.
A. (−1,3)\left( -1,3 \right).
B. (−3,1)\left( -3,1 \right)
C. [−2,2]\left[ -2,2 \right]
D. R−(−2,2)R-\left( -2,2 \right)

Explanation

Solution

For this problem we need to calculate the range of the variable aa if the given equation has a solution. We can observe that the given equation is a quadratic equation in terms of sin⁡x\sin x. We will calculate the solution of the given quadratic equation by using the quadratic formula which is x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}. So, we will first compare the given equation with ax2+bx+ca{{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, 4sin⁡2x+4sin⁡x+a2−3=04{{\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\sin x. Now comparing the above equation with ax2+bx+ca{{x}^{2}}+bx+c, then we will get
a=4a=4, b=4b=4, c=a2−3c={{a}^{2}}-3.
Now the solution of the given equation will be
⇒sin⁡x=−4±42−4(4)(a2−3)2(4) ⇒sin⁡x=−4±16−16a2+488 ⇒sin⁡x=−4±64−16a28 \begin{aligned} & \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{aligned}
Taking 1616 as common from the value 64−16a2\sqrt{64-16{{a}^{2}}}, then we will get
⇒sin⁡x=−4±44−a28\Rightarrow \sin x=\dfrac{-4\pm 4\sqrt{4-{{a}^{2}}}}{8}
To have the value of above solution we need to have 4−a2≥04-{{a}^{2}}\ge 0. Considering this condition and simplifying it, then we will get
⇒4−a2≥0 ⇒22−a2≥0 \begin{aligned} & \Rightarrow 4-{{a}^{2}}\ge 0 \\\ & \Rightarrow {{2}^{2}}-{{a}^{2}}\ge 0 \\\ \end{aligned}
Applying the formula a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) in the above equation, then we will get
⇒(2+a)(2−a)≥0 ⇒2+a≥0 and 2−a≥0 ⇒a≥−2 and a≤2 \begin{aligned} & \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{aligned}
From the above conditions we can write the range of the variable aa as [−2,2]\left[ -2,2 \right].

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