1. ## Trigonometric Equations Help Please!

Please can you tell me how to solve the following equations for values of x between (0.2 π) {π is pie}

a) tan (4x) = 1

b) sin(x-2π) + sin (x - 2π) = ½

c) 2 cos (x + ½) = -1

d) sin x + cos x = 3/2

2. Well when you want to isolate a variable that's inside a trig function, you just take the inverse trig function.

For instance, with the first problem:

$\tan(4x)=1$.

We take $\tan^{-1}$ of both sides (sometimes, $\tan^{-1}$ is written and pronounced $\arctan$). Get:

$\tan^{-1}(\tan(4x)) = \tan^{-1}(1)$, so resolving the left side gives:

$4x = \tan^{-1}(1)$.

We now have to figure out what $\tan^{-1}(1)$ is. Well it's like asking "what angle in a right triangle gives a ratio of opposite side over adjacent side equal to 1?" It should be obvious that a 45-45-90 triangle passes this test, so $\tan^{-1}(1)$ resolves to $45^o$, or $\frac{\pi}{4}$. However, there may be more answers. Notice that $\frac{5\pi}{4}$ works as well. In fact, for any integer n, $\tan\left(n\pi + \frac{\pi}{4}\right) = 1$, so $\tan^{-1}(1)$ could be any of these.

We take this info and go back to our equation:

$4x = \tan^{-1}(1)$. So:

$4x = n\pi + \frac{\pi}{4}$. Then:

$x = \frac{1}{4}\left(n\pi + \frac{\pi}{4}\right)$.

All right, but you were given a restriction of possible values, $0 < x < 2\pi$.

So we want to figure out the only values of n that make our solution valid. $0 < \frac{1}{4}\left(n\pi + \frac{\pi}{4}\right) < 2\pi$,
$0 < n\pi + \frac{\pi}{4} < 8\pi$,

$-\frac{\pi}{4} < n\pi < 8\pi - \frac{\pi}{4}$,

$-\frac{1}{4} < n < 8 - \frac{1}{4}$.

So n could only be 0,1,2,3,4,5,6, or 7.

3. Thanks for your help. I will try to work out the others.

4. How do we known that 45 degrees is pi over 4?

5. Originally Posted by bigfoot
How do we known that 45 degrees is pi over 4?
How does $2\pi"$ relate to $180^{\circ}"$?

Dividing by 8, how then does $\frac{\pi}{4}"$ relate to $45^{\circ}"$?