Question

The equation $k\cos x-3\sin x=k+1$ is solvable only if $k$ belongs to the intarval

• A. $[4,\infty ]$
• B. $[-4,4]$
• C. $(-4,4)$
• D. $[-\infty ,4]$

Answer 1

The correct option is D $[-\infty ,4]$

$k\cos x-3\sin x=k+1$

$\Rightarrow k\cos x-(k+1)=3\sin x$

$\Rightarrow {[k\cos x-(k+1)]}^2={\left(3\sin x\right)}^2$

$\Rightarrow k^2{\cos }^{2}x-2k(k+1)\cos x+{(k+1)}^2=9{\sin }^{2}x$

$\Rightarrow k^2{\cos }^{2}x-(2k^2+2k)\cos x+{(k+1)}^2=9(1-{\cos }^{2}x)$

$\Rightarrow \left(k^2\right){\cos }^{2}x-(2k^2+2k)\cos x+(k^2+2k+1)+9{\cos }^{2}x-9=0$

$\Rightarrow (k^2+9){\cos }^{2}x-(2k^2+2k)\cos x+(k^2+2k-8)=0$

This is the quadratic equation in terms of $\cos x$

$\therefore b^2-4ac\ge 0$

$\Rightarrow {[-(2k^2+2k)]}^2-4(k^2+9)(k^2+2k-8)\ge 0$

$\Rightarrow (4k^4+8k^3+4k^2)-(4k^4+8k^3+4k^2+72k-288)\ge 0$

$\Rightarrow -72k+288\ge 0$

$\Rightarrow 72k-288\le 0$

$\Rightarrow 72k\le 288$

$\Rightarrow k\le 4$

$k$ lies $[-\infty ,4]$

Hence, the answer is $[-\infty ,4].$