Question

How do you find the x intercepts of $y=\sin \left( \dfrac{\pi x}{2} \right)+1$?

Answer 1

We know that the point where the graph meets the x-axis is called the x-intercept. So, from the given equation $y=\sin \left( \dfrac{\pi x}{2} \right)+1$ we should find the point where this equation meets the x-axis. So, we should substitute the value of y is equal to 0, then we should find the respective x values. In this way, we can find the x intercept of $y=\sin \left( \dfrac{\pi x}{2} \right)+1$.

Complete step-by-step solution:

For the given question we are given to solve the x intercepts of $y=\sin \left( \dfrac{\pi x}{2} \right)+1$.
As we know intercepts of any equation will get at $y=0$ , therefore we can find the intercepts of given equation by substituting $y=0$ we will get the intercepts of equation.
Let us consider the above equation as equation (1).
$y=\sin \left( \dfrac{\pi x}{2} \right)+1...............\left( 1 \right)$
Let us substitute $y=0$ in equation (1), we get
$\Rightarrow \sin \left( \dfrac{\pi x}{2} \right)+1=0$
Subtracting with 1 on both sides, we get
$\Rightarrow \sin \left( \dfrac{\pi x}{2} \right)+1-1=-1$
By simplifying a bit, we get
$\Rightarrow \sin \left( \dfrac{\pi x}{2} \right)=-1$
Let us consider the above equation as equation (2).
$\sin \left( \dfrac{\pi x}{2} \right)=-1...........................\left( 2 \right)$
Now for finding the general equation for the equation (2), for that let us consider it as z.
$z=\dfrac{\pi x}{2}$
Therefore, by substituting z in equation (2), we get
$\sin \left( z \right)=-1$
It can be written as
$\sin \left( z \right)=\sin \left( \dfrac{3\pi }{2} \right)$
Let us consider the above equation as equation (3).
$\sin \left( z \right)=\sin \left( \dfrac{3\pi }{2} \right)..................\left( 3 \right)$
As we know the general solution for $\sin x=\sin y$ is $x=n\pi +{{\left( -1 \right)}^{n}}y$. Therefore let us apply it for equation (3).
Therefore, $\dfrac{\pi x}{2}=2n\pi +\left( \dfrac{3\pi }{2} \right)$ where n is an integer.
This happens as in the domain $0 < x < 2\pi$, only for $\sin \left( \dfrac{3\pi }{2} \right)=-1$ and sine ratio has a cycle of $2\pi$.
Now let us consider $\dfrac{\pi x}{2}=2n\pi +\left( \dfrac{3\pi }{2} \right)$ as equation (4).
$\dfrac{\pi x}{2}=2n\pi +\left( \dfrac{3\pi }{2} \right)...........\left( 4 \right)$
Now, by dividing with $\pi$ on both sides of equation (4), we get
$\dfrac{x}{2}=2n+\left( \dfrac{3}{2} \right)$
By multiplying with 2 on both sides, we get
$x=4n+3$
Hence, we have x-intercepts for $y=\sin \left( \dfrac{\pi x}{2} \right)+1$ at $\\{..........,-5,-1,3,7,11,......\\}$.

Note: Students may assume that the general solution for $\sin x=\sin y$ is $x=n\pi +y$ but we know that the general solution for $\sin x=\sin y$ is $x=n\pi +{{\left( -1 \right)}^{n}}y$. If this misconception is followed, then the final answer may get interrupted. Students should also avoid calculation mistakes while solving this problem.