Question

Determine all values of a for which the equation ${\cos }^{4}x-(a+2){\cos }^{2}x-(a+3)=0$ has solutions.

• A. $-1,-2$
• B. $-2,-3$
• C. $-3,-4$
• D. $-2,-4$

Answer 1

The correct option is B $-2,-3$

$\cos {}^{4}x-(a+2)\cos {}^{2}x-(a+3)=0a=-1,\Rightarrow \cos {}^{4}x-\cos {}^{2}x=2$

$\cos {}^{2}x(\cos {}^{2}x-1)=2$

This is not possible.

$a=-2,\Rightarrow \cos {}^{4}x-1=0\cos {}^{4}x=1\cos x=1x=2\pi a=-3,\Rightarrow \cos {}^{4}x+\cos {}^{2}x=0\cos {}^{2}x(\cos {}^{2}x+1)=0\therefore x=\frac{\pi k}{2},x=3\pi ka=-4\cos {}^{4}x+2\cos {}^{2}x+1=0{(\cos {}^{2}x+1)}^2=0$

$\cos {}^{2}x=-1$

This is also not possible.

$\therefore$ $-2,-3$.