Question

The set of values of $k\in R$ such that the equation $\cos 2\theta +\cos \theta +k=0$ admits of a solution for $\theta$ is

• A. $[-3,\frac{3}{2}]$
• B. $[0,+\infty )$
• C. $[-2,0]$
• D. $[0,\frac{9}{4}]$

Answer 1

The correct option is A $[-3,\frac{3}{2}]$

We have, $\cos 2\theta +\cos \theta +k=0$

$\Rightarrow 2{\cos }^2\theta -1+\cos \theta +k=0$

$\Rightarrow k=1-2{\cos }^2\theta -\cos \theta$

$\Rightarrow k=1-2({\cos }^2\theta +\frac{1}{2}\cos \theta +\frac{1}{4}-\frac{1}{4})$

$\Rightarrow k=\frac{3}{2}-2{(\cos \theta +\frac{1}{2})}^2$

we know, $-1\le \cos \theta \le 1$

Thus Range of k for which equation posses solution is,
$k\in [-3,\frac{3}{2}]$