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!
