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Solve the equation

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1) The equation sin(θ) + tan(-θ) = 0

sin(θ) - tan(θ) = 0

sin(θ) - tan(θ) = 0

sin(θ) - [sin(θ)/cos(θ)] = 0

[sin(θ) cos(θ) - sin(θ)]/cos(θ) = 0

sin(θ) cos(θ) - sin(θ) = 0

sin(θ) [cos(θ) - 1] = 0

sin(θ) = 0 , cos(θ) - 1 = 0

Case 1 : sin(θ) = 0

The general solution of sin(θ) = 0, then θ = nπ where n is an integer.

For n = 0,θ = 0

For n = 1,θ = π

For n = 2, θ = 2π

The solutions in the interval [0, 2π) are θ = 0, π.

Case 2 : cos(θ) - 1 = 0

cos(θ) = 1

cos(θ) = cos(0)

Principal value θ = 0

α = 0

The genaral solution of cos(θ) = cos(α) is x = 2nπ ± α, where n is an integer.

If n = 0 then x = 2(0)π ± 0 = 0

If n = 1 then x = 2(1)π ± 0 =

If n = 2 then x = 2(2)π ± 0 = 4π

The solutions in the interval [0, 2π) are θ = 0.

Option B is correct.

 

answered Oct 4, 2014 by david Expert
selected Oct 5, 2014 by tonymate
0 votes

2) The equation 2sin(θ) cos(θ) + cos(θ) = 0

cos(θ) [ 2 sin(θ) + 1] = 0

cos(θ) = 0 and 2 sin(θ) + 1 = 0

Case 1 :

cos(θ) = 0

cos(θ) = cos(π/2)

Principal value θ = π/2.

The genaral solution of cos(θ) = cos(α) is x = 2nπ ± α, where n is an integer.

If n = 0 then x = 2(0)π ± π/2 = π/2, -π/2

If n = 1 then x = 2(1)π ± π/2 = 3π/2, 5π/2

If n = 2 then x = 2(2)π ± π/2 = 4π ± π/2 = 9π/2, 7π/2

The solutions in the interval [0, 2π) are θ = π/2,3π/2.

Case 2 :

2 sin(θ) + 1 = 0

2 sin(θ) = - 1

sin(θ) = - 1/2

The function sin(θ) is negative in third and fourth quadrant.

In third quadrant (π < θ < 3π/2)

sin(θ) = - 1/2 = sin(θ) = - sin(π/6) = sin(π + π/6) = sin(7π/6) ------->θ = 7π/6.

In fourth quadrant (3π/2 < θ < 2π)

sin(θ) = -1/2 = sin(θ) = - sin(π/6) = sin(2π - π/6) = sin(11π/6)-------> θ = 11π/6.

The solutions in the interval [0, 2π) are θ = 7π/6,11π/6.

Solutions are π/2, 3π/2, 7π/6 and 11π/6.

Option C is correct.

answered Oct 4, 2014 by david Expert
edited Oct 4, 2014 by david
0 votes

3) The equation 4 cos(θ) + 1 = 2 cos(θ)

4 cos(θ) - 2cos(θ) = - 1

2 cos(θ) = - 1

cos(θ) =  - 1/2

The function cos(x) is negative in second and third quadrant.

In II quadrant (π/2 < θ < π),

cos(θ) = - 1/2 = cos(θ) = - cos(π/3) = cos(π - π/3) = cos(2π/3) -------> θ = 2π/3.

In III quadrant (π < θ < 3π/2),

cos(θ) = - 1/2 = cos(θ) = - cos(π/3) = cos(π + π/3) = cos(4π/3) -------> θ = 4π/3.

Solutions are 2π/3 and 4π/3.

Option A is correct.

answered Oct 4, 2014 by david Expert

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