Question

The number of all possible triplets $(x,y,z)$ such that $(x+y)+(y+2z)\cos 2\theta +(z-x){\sin }^2\theta =0$ for all $\theta$ is :

• A. $0$
• B. $1$
• C. $3$
• D. Infinite

Answer 1

The correct option is D Infinite

We have ,
$a_1+a_2\cos 2x+a_3{\sin }^{2}x=0\forall x$
$\Rightarrow a_1+a_2\cos 2x+a_3\left(\frac{1-\cos 2x}{2}\right)=0\forall x$
$\Rightarrow (a_1+\frac{1}{2}{a}_3)+(a_2-\frac{1}{2}{a}_3)\cos 2x=0\forall x$
$\Rightarrow a_1+\frac{1}{2}{a}_3=0$ and $a_2-\frac{1}{2}{a}_3=0$
$\Rightarrow a_1=-\frac{K}{2},a_2=\frac{K}{2},a_3=K$ where $KϵR$
Hence, the solutions are $(-\frac{K}{2},\frac{K}{2},K)$ where $K$ is any real number. Thus, the number of triplets is infinite.