Question

If every solution of the equation $3 \cos ^2 x-\cos x-1=0$ is a solution of the equation $a \cos ^2 2 x+b \cos 2 x-1=0$. Then find the value of $(a+b)$.

Answer 1

$1^{\text {st }}$ equation gives,
$\frac{3}{2}(1+\cos 2 x)-1=\cos x \Rightarrow 3(1+\cos 2 x)-2=2 \cos x $
$3 \cos 2 x +1=2 \cos x $
$(3 \cos 2 x +1)^2=4 \cos ^2 x =2(1+\cos 2 x )$
$9 \cos ^2 2 x+4 \cos 2 x-1=0 $ ....(1)
$\text { comparing with } $
$a \cos ^2 2 x + b \cos 2 x -1=0 $ ...(2)
$\frac{a}{9}=\frac{b}{4}=1 \Rightarrow b=4 \text { and } a=9 \Rightarrow a+b=13$