solve the following equations in the given interval.
$\displaystyle \tan(3x+25) = -0.51 $ $\displaystyle -90< x \leq 180 $
I dont know what the addition formula is, Have not been taugh that yet.
heres my working but I dont seem to get the right answers.
$\displaystyle \tan(3x+25) = -0.51 $
$\displaystyle X = 3x+25 $
$\displaystyle tanX = -0.51 $
$\displaystyle tan^{-1} (-0.51) = -27.02 $
$\displaystyle -90< x \leq180 $
$\displaystyle -270< 3x \leq540 $
$\displaystyle -245< 3x+25 \leq 565 $
As tanX is −ve, X is in the 2nd and 4th quadrants
so the solutions between −245<X ≤ 565
should one of the solutions be 27 ?
also;
in the fourth qudrant = 360-27 = 333
second quadrant = 180-27 = 153
but I have to get all the solutions up to 565 , how do I do that ?
The correct answer are; X= −207, −27, 153, 333, 513,
can you explain how they get -27, -207 and 513 from ?