Hello thex10
Welcome to Math Help Forum! Originally Posted by
thex10 Hi!
Having problems getting correct answers for the equation
$\displaystyle sin(pi/3-2x) = sin(x-pi/6)$
Could someone help and point me to the right direction with this one
?
Thanks!
The general solution of the equation:$\displaystyle \sin A = \sin B$
is:$\displaystyle A = n\pi + (-1)^nB, n = 0 , \pm 1, \pm 2, ...$
So for instance,When $\displaystyle n = 0,\; A = B$
When $\displaystyle n = 1,\; A = \pi - B$
When $\displaystyle n = 2,\; A = 2\pi + B$
... and so on
So to solve the equation $\displaystyle \sin(\pi/3-2x) = \sin(x-\pi/6)$
you first write down the general solution, as above, where$\displaystyle A = \pi/3 - 2x$
and $\displaystyle B = x - \pi/6$
which is:$\displaystyle \pi/3 - 2x =n\pi +(-1)^n(x - \pi/6)$
and then see what happens when you take particular values for $\displaystyle n$.
Can you continue from here?
Grandad