Question

For each integer $n>1$, let $S\left(n\right)$ denote the number of solution of the equation $\sin x=\sin nx$ on the interval $[0,\pi ]$ Find the value of $S\left(2\right)+S\left(3\right)+S\left(4\right)$.

• A. 12
• B. 11
• C. 15
• D. 10

Answer 1

The correct option is A 12

$sin\left(x\right)=sin2x$
$sin\left(x\right)=2sin\left(x\right).cos\left(x\right)$
$sin\left(x\right)\left[2cos\right(x)-1]=0$
$sin\left(x\right)=0$ and $cos\left(x\right)=\frac{1}{2}$
$x=0,\pi$ and $x=\frac{\pi }{3}$ ...(i)
Hence there are 3 solutions.
$S\left(3\right)$
$sin\left(x\right)=sin3x$
$sin3x-sinx=0$
$2cos\left(2x\right).sinx=0$
$sin\left(x\right)=0$ and $cos\left(2x\right)=0$
$x=0,\pi$ and $x=\frac{\pi }{4},\frac{3\pi }{4}$
Hence number of solutions are 4.
$S\left(4\right)$
$sin\left(x\right)=sin4x$
$sin4x-sinx=0$
$2cos\left(\frac{5x}{2}\right).sin\left(\frac{3x}{2}\right)=0$
$x=0,\frac{2\pi }{3}$ and $x=\frac{\pi }{5},\frac{3\pi }{5},\pi$
Hence number of solutions are 5.
Thus $S\left(2\right)+S\left(3\right)+S\left(4\right)$
$=3+4+5$
$=12$.