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
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:
$\displaystyle \tan(4x)=1$.
We take $\displaystyle \tan^{-1}$ of both sides (sometimes, $\displaystyle \tan^{-1}$ is written and pronounced $\displaystyle \arctan$). Get:
$\displaystyle \tan^{-1}(\tan(4x)) = \tan^{-1}(1)$, so resolving the left side gives:
$\displaystyle 4x = \tan^{-1}(1)$.
We now have to figure out what $\displaystyle \tan^{-1}(1)$ 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 $\displaystyle \tan^{-1}(1)$ resolves to $\displaystyle 45^o$, or $\displaystyle \frac{\pi}{4}$. However, there may be more answers. Notice that $\displaystyle \frac{5\pi}{4}$ works as well. In fact, for any integer n, $\displaystyle \tan\left(n\pi + \frac{\pi}{4}\right) = 1$, so $\displaystyle \tan^{-1}(1)$ could be any of these.
We take this info and go back to our equation:
$\displaystyle 4x = \tan^{-1}(1)$. So:
$\displaystyle 4x = n\pi + \frac{\pi}{4}$. Then:
$\displaystyle x = \frac{1}{4}\left(n\pi + \frac{\pi}{4}\right)$.
All right, but you were given a restriction of possible values, $\displaystyle 0 < x < 2\pi$.
So we want to figure out the only values of n that make our solution valid. $\displaystyle 0 < \frac{1}{4}\left(n\pi + \frac{\pi}{4}\right) < 2\pi$,
$\displaystyle 0 < n\pi + \frac{\pi}{4} < 8\pi$,
$\displaystyle -\frac{\pi}{4} < n\pi < 8\pi - \frac{\pi}{4}$,
$\displaystyle -\frac{1}{4} < n < 8 - \frac{1}{4}$.
So n could only be 0,1,2,3,4,5,6, or 7.