+895 Solve in the interval \(0\leq \theta<2\pi\).
\(tan(2\theta)+2sin\theta=0\)
\(tan(2\theta)+2cos\theta=0\)
+129899 tan(2θ) + 2sinθ = 0
2tanθ
__________ + 2sinθ = 0
1 - tan^2θ
2tanθ + 2sinθ(1 - tan^2θ) = 0
2tanθ + 2sinθ - 2sinθtan^2θ = 0
tanθ + sinθ - sinθtan^2θ = 0
sinθtan^2θ - tanθ - sinθ = 0
sinθsin^2θ sinθcosθ sinθcos^2θ
_________ - __________ - __________ = 0
cos^2θ cos^2θ cos^2θ
sin^3θ - sinθcosθ - sincos^2θ = 0
sinθ [ sin^2θ - cosθ - cos^2θ] = 0 sinθ = 0 at 0 and pi
1-cos^2θ - cosθ - cos^2θ = 0
-2cos^2θ - cosθ + 1 = 0
2cos^2θ + cosθ - 1 = 0
(2cosθ -1)(cosθ + 1) = 0
2cosθ - 1 = 0 and cosθ + 1 = 0
2cosθ = 1 cosθ = - 1
cosθ = 1/2 And this happens at pi
And this happens at
pi/3 and 5pi/3
So the solutions are θ = 0, pi/3, pi and 5pi/3
Can you do the other one, AT???
