Hi
Can someone help me solve the following equations, coz i have tired and i cannot understand how to do them.
1)sin(2x)=cos(x)
2)sin(8x)=cos(4x)
P.S
1) $\displaystyle \sin{2x} = \cos{x}$
$\displaystyle 2\sin{x}\cos{x} = \cos{x}$
$\displaystyle 2\sin{x}\cos{x} - \cos{x} = 0$
$\displaystyle \cos{x}(2\sin{x} - 1) = 0$
$\displaystyle \cos{x} = 0$ or $\displaystyle 2\sin{x} - 1 = 0$.
Case 1: $\displaystyle \cos{x} = 0$
$\displaystyle x = \left\{\frac{\pi}{2}, \frac{3\pi}{2}\right\} + 2\pi n$, where $\displaystyle n$ is an integer representing the number of times you have gone around the unit circle...
Case 2: $\displaystyle 2\sin{x} - 1 = 0$
$\displaystyle 2\sin{x} = 1$
$\displaystyle \sin{x} = \frac{1}{2}$
$\displaystyle x = \left\{\frac{\pi}{6}, \frac{5\pi}{6}\right\} + 2\pi n$.
So $\displaystyle x = \left\{\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}\right\} + 2\pi n$.
2) $\displaystyle \sin{8x} = \cos{4x}$
$\displaystyle 2\sin{4x}\cos{4x} = \cos{4x}$
Note that if we let $\displaystyle 4x = X$ we have the exact same equation as in Q.1.
So
$\displaystyle 4x = \left\{\frac{\pi}{6},\frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}\right\} + 2\pi n$
$\displaystyle x = \left\{ \frac{\pi}{24}, \frac{\pi}{8}, \frac{5\pi}{24}, \frac{3\pi}{8} \right\} + \frac{\pi n}{2}$