Can you solve #sin^2theta+sintheta-1=0# by factoring or quadratic formula?

1 Answer
Feb 28, 2017

#38^@17 + k360^@#
#141^@83 + k360^@#

Explanation:

#sin^2 t + sin t - 1 = 0#
Solve this quadratic equation for sin t by using the improved quadratic formula (Socratic Search).
#D = d^2 = b^ - 4ac = 1 + 4 = 5# --> #d = +- sqrt5#
There are 2 real roots:
#sin t = -b/(2a) +- d/(2a) = - 1/2 +- sqrt5/2 = (-1 +- sqrt5)/2#
a. #sin t = (- 1 - sqrt5)/2 = -3.236/2 = - 1.62# (rejected as < - 1)
b. #sin t = (-1 + sqrt5)/2 = (1.236)/2 = 0.618#
Calculator gives -->
sin t = 0.618 --> #t = 38^@ 17 + k360^@#
Unit circle gives another t that has the same sine value (0.618):
#x = 180 - 38.17 = 141^@83 + k360^@#