#sin^4x - 5(1 - sin^2 x) + 1 = 0#
#sin^4 x - 5 + 5sin^2 x + 1= 0#
#sin^4 x + 5sin^2 x - 4 = 0#
Call #sin^2 x = t#, we get a quadratic equation in t.
#t^2 + 5t - 4 = 0#
#D = d^2 = b^2 - 4ac = 25 + 16 = 41# --> #d= +- sqrt41#
There are 2 real roots:
#t = - 5/2 +- sqrt41/2 = (-5 +- sqrt41)/2#
a. #t = sin^2 x = (- 5 + sqrt41)/2 = 0.70#
#sin x = +- 0.838#
Unit circle and calculator give -->
1. #sin x = 0.838# -->
#x = 56^@93#, and #x = 180 - 56^@93 = 123^@07#
2. sin x = - 0.838 -->
#x = - 56^@93#, or #x = 360 - 56.93 = 303^@07# (co-terminal), and
#x = 180 - (- 56.93) = 180 + 56.93 = 236^@93#
b. #sin^2 x = t = - 5/2 - sqrt41/2 = (-5 - sqrt41)/2#. (Rejected)
For general answers, add #k360^@#
