Question

If $\cos (4y-3x-2)-\cos (4y+3x+2)=2+2\ln (k^4-255)$ and $\cos (4y-3x-2)+\cos (4y+3x+2)=2k+8$ have real solutions $(x,y)$, then the range of $k$ is

• A. $[3,4]$
• B. $[-3,4]$
• C. $[-4,-3]$
• D. $[-4,3]$

Answer 1

The correct option is C $[-4,-3]$

$\cos (4y-3x-2)-\cos (4y+3x+2)=2+2\ln (k^4-255)\cos (4y-3x-2)+\cos (4y+3x+2)=2k+8$
Assuming $4y=A,3x+2=B$, we get
$\cos (A-B)-\cos (A+B)=2(1+\ln (k^4-255\left)\right)\Rightarrow 2\sin A\sin B=2(1+\ln (k^4-255\left)\right)\Rightarrow 1+\ln (k^4-255)=\sin A\sin B\Rightarrow 1+\ln (k^4-255)\le 1\Rightarrow \ln (k^4-255)\le 0\Rightarrow k^4-255\le 1\Rightarrow k^4\le 256\Rightarrow k\in [-4,4]\cdots \left(1\right)$

Also,
$\cos (A-B)+\cos (A+B)=2k+8\Rightarrow 2\cos A\cos B=2(k+4)\Rightarrow k+4=\cos A\cos B\Rightarrow k+4\le 1\Rightarrow k\le -3\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get
$k\in [-4,-3]$