Question

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

Answer 1

Hint: 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.