Question

The value of x which satisfy the equation
a$sin(x+1)+bsin(x+2)+csin(x+3)=0$ given that the coefficient a, b, c are chosen so that there are atleast 2 solution of this equation in the interval (0,$\pi$), is

• A. $x=\frac{x}{6}$
• B. $x=\frac{x}{3}$
  
• C. $x=\frac{x}{2}$
  
• D. $x=\frac{2x}{3}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $x=\frac{x}{6}$
B $x=\frac{x}{3}$

C $x=\frac{x}{2}$

D $x=\frac{2x}{3}$

Using $sin(x+\alpha )=sinxcos\alpha +cos\alpha sin\alpha +for\alpha =1,2,3$

We get $asin(x+1)+bsin(x+2)+csin(x+3)=\lambda cos(x+\varphi )$ for same $\lambda \&\varphi$

If $\lambda \ne 0\Rightarrow cos(x+\varphi )=0$

$\Rightarrow x+\varphi =(2n+1{)}_{\frac{\pi }{2}},nϵz\Rightarrow x=(2n+1{)}_{\frac{\pi }{2}}-\varphi$

$\because$ Two consective roots differ by $\pi$

$\therefore$ The equation cannot have two or more roots in (0,$\pi$) so we must have $\lambda =0$. In this case equation is satisfied for all $xϵR$.