Find roots of the equation $y=2 \sin(3x+40), \;x \in (-2\pi, 2\pi)$

Question

Find the roots of the equation:

$$y=2 \sin(3x+40), \;\;x \in (-2\pi, 2\pi)$$

In the book given that there are 12 roots exist. I am able get only 2 roots. Could anyone explain?

Answer 1

$$y = 2\sin(3x+40)$$

Set $y = 0$. $$0 = 2\sin(3x+40) \implies \sin(3x+40) = 0 \implies 3x+40 = \sin^{-1}0$$

From here, for all $n \in \mathbb{Z}$: $$\sin^{-1}0 = 0+2\pi n \text{ OR } \sin^{-1}0 = \pi+2\pi n \\ \implies\sin^{-1}0 = \pi n$$

Solve the equation now. $$3x+40 = \pi n \implies 3x = \pi n-40 \implies \boxed{x = \frac{\pi n-40}{3} = n\frac{\pi}{3}-\frac{40}{3}}$$ From here, we can find the number of roots for the restricting domain of $x \in (-2\pi, 2\pi)$. $$-2\pi < \frac{\pi n-40}{3} < 2\pi$$ $$-6\pi < -\pi n-40 < 6\pi$$ $$-6\pi+40< -\pi n < 6\pi+40$$ $$\frac{-6\pi+40}{-\pi}> n > \frac{6\pi+40}{-\pi}$$ $$6-{\frac{40}{\pi}} > n > -6-\frac{40}{\pi}$$

Subtract the maximum and minimum values. $$6-\frac{40}{\pi}-\biggr(-6-\frac{40}{\pi}\biggr) = 12$$ So, there are $12$ roots. As a matter of fact, the answer would remain $12$ regardless of the value of $h$ in $y = 2\sin(3x-h)$.

Another way of thinking about it would be to consider the periodic nature of sine. For $y = a\sin[b(x-h)]+k$, the length of a period is given by $\frac{2\pi}{b}$.

$$\frac{2\pi}{b} = \frac{2\pi}{3}$$

For $x \in (-2\pi, 2\pi)$, the number of cycles/periods can be found by $$\frac{4\pi}{\frac{2\pi}{3}} = 6$$ Each period has $2$ roots, so there’ll be $12$ roots for the given function.