How do you find the exact value of #(2sinthetacostheta)^2+4sin2theta-1=0# in the interval #0<=theta<2pi#?
1 Answer
Dec 11, 2016
We can rewrite
#=>(sin2theta)^2 + 4sin2theta -1=0#
We let
#=>t^2 + 4t - 1 = 0#
#=>t= (-4 +- sqrt(4^2 - 4 xx 1 xx -1))/(2 xx 1)#
#=>t = (-4 +- sqrt(20))/2#
#=>t = (-4 +- 2sqrt(5))/2#
#=>t = -2 +- sqrt(5)#
#:.sin2theta = -2 +- sqrt(5)#
#=>2theta = arcsin(-2 +- sqrt(5))#
#=>theta = 1/2arcsin(-2 +-sqrt(5))#
Hopefully this helps!
