Why Does Sin(x) = 1 Have Different Solutions Compared to Sin(2x) = 1?

[danglade]

Suppose you are solving equations in the interval 0<=xx<=2pi...Without actually solving equations, what is the difference between the number of solutions of sinx=1 and sin2x=1? How do you account for this difference

 

[Gib Z]

Well since sin 2x = 2 sin x cos x = 1, and you also want solutions to sin x =1, we can see that the number of solutions of sin x = sin 2x = 1 is the number of times cos x is equal to 1 in the interval.

 

[rohanprabhu]

sin x = 1
sin x = sin (\frac{\pi}{2})
x = n\pi + (-1)^n (\frac{\pi}{2})

But, as x \in [0, 2\pi]

Hence, select the values for n (n \in N), such that x \in [0, 2\pi].

The satisfying values are: n \in \{0, 1\} Put this values for x, and you shall get x \in \{\frac{\pi}{2}\}. This is because when we have \alpha as \frac{\pi}{2}, we get the same solutions for n = 0; n = 1.

Do the same for 2x (x \in [0, 2\pi]), (2x \in [0, 4\pi]) and you shall get the solutions for 2x. It again gives us:

2x = n\pi + (-1)^n (\frac{\pi}{2})

Here, {0, 1, 2, 3} satisfy 'n', giving 2 unique solutions i.e. x \in \{\frac{\pi}{4}, \frac{5\pi}{4}\}.