# Thread: more solving trig questions

1. ## more solving trig questions

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

2. Originally Posted by Paymemoney
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) $\sin{2x} = \cos{x}$

$2\sin{x}\cos{x} = \cos{x}$

$2\sin{x}\cos{x} - \cos{x} = 0$

$\cos{x}(2\sin{x} - 1) = 0$

$\cos{x} = 0$ or $2\sin{x} - 1 = 0$.

Case 1: $\cos{x} = 0$

$x = \left\{\frac{\pi}{2}, \frac{3\pi}{2}\right\} + 2\pi n$, where $n$ is an integer representing the number of times you have gone around the unit circle...

Case 2: $2\sin{x} - 1 = 0$

$2\sin{x} = 1$

$\sin{x} = \frac{1}{2}$

$x = \left\{\frac{\pi}{6}, \frac{5\pi}{6}\right\} + 2\pi n$.

So $x = \left\{\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}\right\} + 2\pi n$.

3. Originally Posted by Paymemoney
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
2) $\sin{8x} = \cos{4x}$

$2\sin{4x}\cos{4x} = \cos{4x}$

Note that if we let $4x = X$ we have the exact same equation as in Q.1.

So

$4x = \left\{\frac{\pi}{6},\frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}\right\} + 2\pi n$

$x = \left\{ \frac{\pi}{24}, \frac{\pi}{8}, \frac{5\pi}{24}, \frac{3\pi}{8} \right\} + \frac{\pi n}{2}$