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Solve in the interval \(0\leq \theta<2\pi\).

 

\(tan(2\theta)+2sin\theta=0\)

 

\(tan(2\theta)+2cos\theta=0\)

 May 13, 2019
 #1
avatar+101084 
+1

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???

 

 

cool cool cool

 May 14, 2019
 #2
avatar+881 
+1

Yes, thanks.

AdamTaurus  May 14, 2019

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