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.
$$\left[ {0,\frac{9}{8}} \right]$$
B.
$$\left[ {0, + \infty } \right)$$
C.
$$\left[ { - 2,0} \right]$$
D.
None of these
Answer :
$$\left[ {0,\frac{9}{8}} \right]$$
Solution :
$$\eqalign{
& 2{\cos ^2}\theta + \cos \theta + \left( {k - 1} \right) = 0 \cr
& \therefore \,\,\cos \theta = \frac{{ - 1 \pm \sqrt {1 - 8\left( {k - 1} \right)} }}{4} = \frac{{ - 1 \pm \sqrt {9 - 8k} }}{4}. \cr} $$
For real, $$\cos\theta ,9 - 8k \geqslant 0,\,{\text{i}}{\text{.e}}{\text{., }}k \leqslant \frac{9}{8}.$$
Also, $$ - 1 \leqslant \frac{{ - 1 \pm \sqrt {9 - 8k} }}{4} \leqslant 1\,\,{\text{or, }} - 4 \leqslant - 1 \pm \sqrt {9 - 8k} \leqslant 4$$
or, $$ - 3 \leqslant \pm \sqrt {9 - 8k} \leqslant 5$$
$$\eqalign{
& \therefore \,\, - 3 \leqslant - \sqrt {9 - 8k} \,\,{\text{and }}\sqrt {9 - 8k} \leqslant 5 \cr
& \Rightarrow \,\,9 - 8k \leqslant 9\,\,{\text{and }}9 - 8k \leqslant 25 \cr
& \therefore \,\,k \geqslant 0\,\,{\text{and }}k \geqslant - 2 \cr
& \Rightarrow \,\,k \geqslant 0. \cr} $$