Solve this quadratic equation for sin t.
3sin^2 t + 10 sin t + 2 = 03sin2t+10sint+2=0
D = b^2 - 4ac = 100 - 24 = 76D=b2−4ac=100−24=76 --> d = +- 2sqrt19d=±2√19
There are 2 real roots:
sin t = - b/(2a )+- d/(2a) = - 10/6 +- (2sqrt19)/6 = (-5 +- sqrt19)/3sint=−b2a±d2a=−106±2√196=−5±√193
The root sin t = (-5 - sqrt19)/3 = - 3.12sint=−5−√193=−3.12 is rejected as < -1.
The root sin t = (-5 + sqrt19)/3 = - 0.213sint=−5+√193=−0.213 .
Calculator and the unit circle give 2 solution arcs:
t1 = -12^@34t1=−12∘34 , or t1 = 347^@66t1=347∘66 (co-terminal), and
t2 = 180 - (-12.34) = 180 + 12.34 = 192^@34t2=180−(−12.34)=180+12.34=192∘34
Answers for (0, 2pi):(0,2π):
192^@34 and 347^@66192∘34and347∘66
For general answers, add k360^@k360∘
