Question

Sinx= 2x/(21π) find no of roots

Answer 1

21

Answer 2

The problem asks us to find the number of roots for the equation $\sin x = \frac{2x}{21\pi}$. This is a transcendental equation, best solved by graphical analysis. Let $f(x) = \sin x$ and $g(x) = \frac{2x}{21\pi}$. We need to find the number of intersections of the curve $y = \sin x$ and the line $y = \frac{2x}{21\pi}$.

1. Analyze $y = \sin x$: This is an oscillatory function with a range of $[-1, 1]$. It passes through the origin $(0,0)$ and is an odd function ($\sin(-x) = -\sin x$).

2. Analyze $y = \frac{2x}{21\pi}$: This is a straight line passing through the origin $(0,0)$. The slope is $m = \frac{2}{21\pi}$, which is a small positive value (approximately $2/(21 \times 3.14159) \approx 0.03$). Since it's a line with a positive slope passing through the origin, it's an odd function ($g(-x) = -g(x)$).

3. Check for $x=0$: For $x=0$, $\sin(0) = 0$ and $\frac{2(0)}{21\pi} = 0$. So, $x=0$ is one root.

4. Determine the range of possible roots: Since the range of $\sin x$ is $[-1, 1]$, for an intersection to occur, we must have $-1 \le \frac{2x}{21\pi} \le 1$. This implies: $-1 \le \frac{2x}{21\pi} \implies -21\pi \le 2x \implies x \ge -\frac{21\pi}{2}$ $\frac{2x}{21\pi} \le 1 \implies 2x \le 21\pi \implies x \le \frac{21\pi}{2}$ So, all roots must lie in the interval $\left[-\frac{21\pi}{2}, \frac{21\pi}{2}\right]$. Note that $\frac{21\pi}{2} = 10.5\pi$.

5. Analyze roots for $x > 0$: The line $y = \frac{2x}{21\pi}$ starts at $(0,0)$ and increases linearly. At $x = \frac{21\pi}{2}$, the value of the line is $y = \frac{2}{21\pi} \times \frac{21\pi}{2} = 1$. So, the line connects $(0,0)$ to $(\frac{21\pi}{2}, 1)$.

   Let's examine the intervals where $\sin x$ is positive or negative. The line $y = \frac{2x}{21\pi}$ is always positive for $x > 0$. Therefore, there will be no intersections when $\sin x < 0$. This means we only need to consider intervals where $\sin x \ge 0$, i.e., intervals of the form $[2k\pi, (2k+1)\pi]$ for integer $k \ge 0$.

   Let's find the number of such intervals within $(0, 10.5\pi]$. The intervals are:

   • $(0, \pi]$: $\sin x$ goes from $0$ to $1$ then back to $0$. The line goes from $0$ to $\frac{2\pi}{21\pi} = \frac{2}{21}$. At $x=0$, slopes are $\cos(0)=1$ for $\sin x$ and $\frac{2}{21\pi} \approx 0.03$ for the line. Since $1 > 0.03$, $\sin x$ rises faster than the line initially. At $x=\pi$, $\sin(\pi)=0$ and $y_{line} = \frac{2}{21} > 0$. So the line is above $\sin x$ at $x=\pi$. Since $\sin x$ starts above the line and ends below the line in $(0, \pi)$, there must be one intersection in $(0, \pi)$. (Let's call this $x_1$).

   • $(\pi, 2\pi]$: $\sin x < 0$. No intersections.

   • $(2\pi, 3\pi]$: $\sin x > 0$. The line goes from $\frac{4\pi}{21\pi} = \frac{4}{21}$ to $\frac{6\pi}{21\pi} = \frac{6}{21}$. At $x=2\pi$, $\sin(2\pi)=0$, $y_{line} = \frac{4}{21}$. Line is above $\sin x$. At $x=2\pi+\pi/2 = 5\pi/2$, $\sin(5\pi/2)=1$, $y_{line} = \frac{2(5\pi/2)}{21\pi} = \frac{5}{21}$. Since $\frac{5}{21} < 1$, the line is below the peak of $\sin x$. At $x=3\pi$, $\sin(3\pi)=0$, $y_{line} = \frac{6}{21}$. Line is above $\sin x$. Since the line starts above $\sin x$, goes below the peak of $\sin x$, and ends above $\sin x$, there will be two intersections in $(2\pi, 3\pi)$. (Let's call them $x_2, x_3$).

   This pattern of two intersections in each interval $(2k\pi, (2k+1)\pi)$ will continue as long as the line $y=\frac{2x}{21\pi}$ is below 1 at $x=(2k+1/2)\pi$ and above 0 at $x=(2k+1)\pi$. The line is always above 0 for $x>0$. We need to check when $\frac{2x}{21\pi} \le 1$. This is $x \le \frac{21\pi}{2} = 10.5\pi$.

   Let's list the intervals $(2k\pi, (2k+1)\pi)$ up to $10.5\pi$:

   • $(0, \pi)$: 1 root
   • $(2\pi, 3\pi)$: 2 roots
   • $(4\pi, 5\pi)$: 2 roots
   • $(6\pi, 7\pi)$: 2 roots
   • $(8\pi, 9\pi)$: 2 roots
   • $(10\pi, 10.5\pi]$: $\sin x$ is positive in this range. At $x=10\pi$, $\sin(10\pi)=0$, $y_{line} = \frac{2(10\pi)}{21\pi} = \frac{20}{21}$. Line is above $\sin x$. At $x=10\pi+\pi/2 = 10.5\pi$, $\sin(10.5\pi) = \sin(10\pi+\pi/2) = \sin(\pi/2) = 1$. At $x=10.5\pi$, $y_{line} = \frac{2(10.5\pi)}{21\pi} = \frac{21\pi}{21\pi} = 1$. So, at $x=10.5\pi$, the line touches the peak of the sine wave. This is an intersection point. Since the line starts above $\sin x$ at $10\pi$ and touches $\sin x$ at $10.5\pi$, there is exactly one intersection in $(10\pi, 10.5\pi]$. (Let's call this $x_{10}$).

   Total positive roots: $1 + 2 + 2 + 2 + 2 + 1 = 10$ roots.

6. Analyze roots for $x < 0$: Since both $f(x)=\sin x$ and $g(x)=\frac{2x}{21\pi}$ are odd functions, if $x_0$ is a non-zero root, then $-x_0$ must also be a root. We found 10 positive roots. Therefore, there must be 10 negative roots.

7. Total number of roots:

   • One root at $x=0$.
   • Ten positive roots.
   • Ten negative roots. Total roots = $1 + 10 + 10 = 21$.

The final answer is $\boxed{\text{21}}$.

Explanation of the solution:

1. Represent the equation as the intersection of two functions: $y = \sin x$ and $y = \frac{2x}{21\pi}$.
2. Observe that both functions pass through the origin $(0,0)$, so $x=0$ is one root.
3. Determine the range of $x$ for which roots can exist: Since $-1 \le \sin x \le 1$, we must have $-1 \le \frac{2x}{21\pi} \le 1$, which simplifies to $-\frac{21\pi}{2} \le x \le \frac{21\pi}{2}$. Note that $\frac{21\pi}{2} = 10.5\pi$.
4. Analyze positive roots ($x > 0$):
   • The line $y = \frac{2x}{21\pi}$ is always positive for $x > 0$. Therefore, it cannot intersect $y = \sin x$ when $\sin x < 0$ (i.e., in intervals like $(\pi, 2\pi), (3\pi, 4\pi)$, etc.).
   • We only need to consider intervals where $\sin x \ge 0$, which are of the form $[2k\pi, (2k+1)\pi]$.
   • In the interval $(0, \pi)$: $\sin x$ goes from $0$ to $1$ and back to $0$. The line goes from $0$ to $\frac{2}{21}$. Since the slope of $\sin x$ at $x=0$ is $1$ (greater than the line's slope $\frac{2}{21\pi}$), $\sin x$ starts above the line. At $x=\pi$, $\sin x = 0$ while the line is at $\frac{2}{21} > 0$. Thus, there is one intersection in $(0, \pi)$.
   • For intervals $(2k\pi, (2k+1)\pi)$ where $k \ge 1$: The line starts above $\sin x$ at $2k\pi$ (since $\sin(2k\pi)=0$ and line value is positive). At the peak of $\sin x$ (at $x=2k\pi+\pi/2$), $\sin x = 1$. The line value is $\frac{2(2k\pi+\pi/2)}{21\pi} = \frac{4k+1}{21}$. As long as $\frac{4k+1}{21} < 1$, the line is below the peak. At $x=(2k+1)\pi$, $\sin x = 0$ and the line is at $\frac{2(2k+1)\pi}{21\pi} = \frac{4k+2}{21} > 0$. This configuration implies two intersections in each such interval.
   • Let's check the condition $\frac{4k+1}{21} < 1$: $4k+1 < 21 \implies 4k < 20 \implies k < 5$. So, for $k=1, 2, 3, 4$, there are two roots in each interval: $(2\pi, 3\pi)$: 2 roots $(4\pi, 5\pi)$: 2 roots $(6\pi, 7\pi)$: 2 roots $(8\pi, 9\pi)$: 2 roots
   • Consider the last relevant interval $(10\pi, 10.5\pi]$: At $x=10\pi$, $\sin(10\pi)=0$, line value is $\frac{20}{21}$. Line is above $\sin x$. At $x=10.5\pi$, $\sin(10.5\pi)=1$, line value is $\frac{2(10.5\pi)}{21\pi}=1$. The line touches the sine curve at its peak. This implies exactly one intersection in $(10\pi, 10.5\pi]$.
   • Total positive roots = $1$ (from $(0, \pi)$) + $2 \times 4$ (from $k=1,2,3,4$) + $1$ (from $(10\pi, 10.5\pi]$) = $1 + 8 + 1 = 10$.
5. Analyze negative roots ($x < 0$): Since both $y=\sin x$ and $y=\frac{2x}{21\pi}$ are odd functions, if $x_0$ is a non-zero root, then $-x_0$ is also a root. Since there are 10 positive roots, there are 10 negative roots.
6. Total roots = $1$ (for $x=0$) + $10$ (positive roots) + $10$ (negative roots) = $21$.