Solve on the interval of 0<or equal to. (theta) < 2(pi).

Question

a). -cos(theta)= (1/2) b). 4sin(theta)+3sqrt.3 = sqrt.3 c). cos(theta)= -sqrt. 3/2? 

 

 

Answer 1

(a)

-cosθ = ½

cosθ =- ½

θ = 2π/3

General solution = 2nπ ± θ where n = 0,1,2,...   and 0<θ<π

θ = 2nπ±2π/3

For n = 0

θ = ± 2π/3

Here -2π/3<0.So it is not valid.

Here 2π/3<2π.So it is valid.

For n = 1

θ = 2π± 2π/3

θ = 2π+2π/3 , 2π- 2π/3

θ = 8π/3 , 4π/3

Here 4π/3<2π.So it is valid.

Here 8π/3>2π.So it is not valid.

The Solution is  { 2π/3 , 4π/3 }

Answer 2

(c)

cosθ = -√3/2

θ = 5π/6

General solution = 2nπ ± θ where n = 0,1,2,...   and 0<θ<π

θ = 2nπ±5π/6

For n = 0

θ = ± 5π/6

Here -5π/6<0.So it is not valid.

Here 5π/6<2π.So it is valid.

For n = 1

θ = 2π± 5π/6

θ = 2π+5π/6 , 2π - 5π/6

θ = 17π/6 , 7π/6

Here 7π/6<2π.So it is valid.

Here 17π/6>2π.So it is not valid.

The Solution is  { 5π/6 , 7π/6 }

Answer 3

(b)

4sinθ + 3√3 = √3

4sinθ = - 3√3 + √3

4sinθ = - 2√3

sinθ = - 2√3 / 4

sinθ = - √3 / 2

θ = - π/3

General solution = nπ + (-1)ⁿθ where n = 0,1,2,...   and 0<θ<2π

θ = nπ + (-1)ⁿ(-π/3)

θ = nπ - (-1)ⁿ(π/3)

For n = 0

θ = -π/3

Here -π/3<0.So it is not valid.

For n = 1

θ = π - (-1)π/3

θ = π + π/3

θ = 4π/3

Here -2π/3<2π.So it is valid.

For n = 2

θ = 2π - (-1)²π/3

θ = 2π - π/3

θ = 5π/3

Here 5π/3<π.So it is valid.

For n = 3

θ = 3π - (-1)³π/3

θ = 3π + π/3

θ = 10π/3

Here 10π/3>2π.So it is not valid.

And for all odd numbers it is not valid.

For n = 4

θ = 4π - (-1)⁴π/3

θ = 4π - π/3

θ = 11π/3

Here 11π/3>2π.So it is not valid.

And for all even numbers greater than n = 4 , it is not valid.

The Solution is  { 4π/3 , 5π/3 }