Given that A is measured in radians, find all the values of A between 0 and 5 for which 5sin(2A+3) + 3cos(2A+3)=0
Let X = (2A +3)
So,
5sinX +3cosX = 0
5sinX = -3cosX
sinX /cosX = -3/5
tanX = -3/5 ----------negative tan value, so X is in the 2nd or 4th quadrants
X = arctan(-3/5) = -0.5404195 radian
In the 2nd quadrant,
X = pi -0.5404195 = 2.6012 rad
2A +3 = 2.6012
A = (2.6012 -3) /2 = -0.1994 rad less than 0, so not counted.
In the 4th quadrant,
X = 2pi -0.5404195 = 5.7428 rad
2A +3 = 5.7428
A = (5.7428 -3) /2 = 1.3714 rad -------one counted.
Now, the period of the tangent function is pi.
Meaning, for every pi radians, the X here will repeat having tangent value of -0.5404195.
So, in the 2nd quadrant, after one revolution,
X = 3pi -0.5404195 = 8.8844 rad
A = (8.8844 -3) /2 = 2.9422 rad ------two counted.
In the next 4th quadrant,
X = 4pi -0.5404195 = 12.0260
A = (12.0260 -3)/2 = 4.513 rad --------3 counted.
In the next 2nd quadrant again,
X = 5pi -0.5404195 = 15.1675 rad
A = (15.1675 -3) /2 = 6.0838 rad ------more than 5.0, so not counted anymore.
Therefore, A = 1.3714, 2.9422 and 4.513 radians. ----------answer.