Trigonometric Equations Help Please!

**bigfoot** · March 24th 2009, 01:19 PM

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

**gosualite** · March 24th 2009, 01:41 PM

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.