How do you solve #3sin^2(t)+ 10sin(t)+ 2 = 0#?

1 Answer
Nghi N.
Oct 10, 2016

#192^@34, 347^@66#

Explanation:

Solve this quadratic equation for sin t.
#3sin^2 t + 10 sin t + 2 = 0#
#D = b^2 - 4ac = 100 - 24 = 76# --> #d = +- 2sqrt19#
There are 2 real roots:
#sin t = - b/(2a )+- d/(2a) = - 10/6 +- (2sqrt19)/6 = (-5 +- sqrt19)/3#
The root #sin t = (-5 - sqrt19)/3 = - 3.12# is rejected as < -1.
The root #sin t = (-5 + sqrt19)/3 = - 0.213# .
Calculator and the unit circle give 2 solution arcs:
#t1 = -12^@34# , or #t1 = 347^@66# (co-terminal), and
#t2 = 180 - (-12.34) = 180 + 12.34 = 192^@34#

Answers for #(0, 2pi):#
#192^@34 and 347^@66#
For general answers, add #k360^@#

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