Question

$\cos 2x+7=a(2-\sin x)$ can have a real solution for

• A. All values of $a$
• B. $a\in \left[2,6\right]$
• C. $a\in \left(-\infty ,2\right)$
• D. $a\in \left(0,\infty \right)$

Answer 1

The correct option is B

$a\in \left[2,6\right]$

Explanation for the correct option:

Step 1. find the solution of given equation:

$\cos 2x+a\sin x=2a-7$

$\Rightarrow$ $1-2{\sin }^{2}x+a\sin x=2a-7$ ; $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{-}\mathbf{2}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{x}\right]$

$\Rightarrow 2{\sin }^{2}x-a\sin x+2a-8=0$

Step 2. Find the roots of above equation:

$\Rightarrow$$\sin x=\frac{a\pm \sqrt{a^2-16(a-4)}}{4}$ ; $\left[\mathbf{\because }\mathit{x}\mathbf{=}\frac{\mathbf{-}\mathbf{b}\mathbf{\pm }\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}\right]$

​​$=\frac{a\pm (a-8)}{4}$

$\Rightarrow$$\sin x=\frac{a-4}{2}or\sin x=2(notpossible)$

$\Rightarrow \sin x=\frac{a-4}{2}$

$\because -1\le \sin x\le 1$

$\therefore -1\le \frac{a-4}{2}\le 1$

$\Rightarrow 2\le a\le 6$

$⟹\mathit{a}\mathbf{\in }\mathbf{[}\mathbf{2}\mathbf{,}\mathbf{6}\mathbf{]}$

Hence, option ‘B’ is Correct.