Solving equations

1. Jan 8, 2008

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

2. Jan 8, 2008

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.

3. Jan 8, 2008

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. $x \in \{\frac{\pi}{4}, \frac{5\pi}{4}\}$.

Last edited: Jan 9, 2008