Question

$\cos 2x+k\sin x=2k-7$ has a solution for

• A. $k>2$
• B. $k<2$
• C. $k>3$
• D. $2\le k\le 6$

Answer 1

The correct option is D

$2\le k\le 6$

Explanation for the correct option:

Step 1. Solve the given equation:

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

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

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

Step 2. Find the roots of above equation:

$\Rightarrow$$\sin x=\frac{+k\pm \sqrt{k^2-8(2k-8)}}{4}$ ; $\left[\mathbf{\because }\mathit{x}\mathbf{=}\frac{\mathbf{-}\mathbf{b}\mathbf{\pm }\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}\right]$

$=\frac{k\pm (k-8)}{4}$
$\Rightarrow$$\sin x=\frac{k-4}{2},\sin x=2$

As we know, the range of $\sin x$is $\left[-1,1\right]$ so it cannot be greater than $1$

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

$\Rightarrow -1\le \sin x\le 1$

$\Rightarrow -1\le \frac{k-4}{2}\le 1$

$\Rightarrow -2\le k-4\le 2$

$\Rightarrow \mathbf{2}\mathbf{\le }\mathit{k}\mathbf{\le }\mathbf{6}$

Hence, Option ‘D’ is Correct.