How do you solve 3 csc^2 5x=4 ?

Question

trigonometry question.

Answer 1

The trigonometric equation is 3 csc² (5x) = 4.

csc² (5x) = 4/3

csc (5x) = ±(√4/3)

csc (5x) = ±(2/√3)

Using the reciprocal identity csc (x) = 1/sin (x).

1/sin (5x) = ±(2/√3)

sin (5x) = 1/±(2/√3)

⇒ sin (5x) = ±(√3/2).

sin(5x) = √3/2 and sin(5x) = - √3/2.

Let 5x = t.

sin(t) = √3/2 and sin(t) = - √3/2.

The function sin(t) has a period of 2π, first find all solutions in the interval [0, 2π).

• sin(t) = √3/2.

The function sin(t) is positive in first and second quadrant.

sin(t) = √3/2 ⇒ t = sin^{- 1}(√3/2).

In first Quadrant, 0 ≤ t ≤ π/2.

t= sin^{- 1}(√3/2) = sin^{- 1}(sin(π/3)) = sin^{- 1}(sin (0 + π/3)) = sin^{- 1}(sin π/3) = π/3.

In second Quadrant, π/2 ≤ t ≤ π.

t = sin^{- 1}(√3/2) = sin^{- 1}(sin (π/3)) = sin^{- 1}(sin ( π - π/3)) = sin^{- 1}(sin 2π/3) = 2π/3.

Finally, add multiples of 2π to each of these solutions to get the general form t = 2nπ + π/3 and t = 2nπ + 2π/3

where n is integer.

Put t = 5x ⇒ 5x = 2nπ + π/3 and 5x = 2nπ + 2π/3

⇒ x = (2nπ + π/3)/5 and x = (2nπ + 2π/3)/5, where n ∈ Z.

Answer 2

Contd.......

• sin(t) = - √3/2.

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

sin(t) = √3/2 ⇒ t = sin^{- 1}(- √3/2).

In third Quadrant,  π ≤ t ≤ 3π/2.

t = sin^{- 1}(- √3/2) = sin^{- 1}(sin (-π/3)) = sin^{- 1}(sin (π + π/3)) = sin^{- 1}(sin 4π/3) = 4π/3.

In fourth Quadrant, 3π/2 ≤ t ≤ 2π.

t= sin^{- 1}(- √3/2)) = sin^{- 1}(sin (- π/3)) = sin^{- 1}(sin ( 2π - π/3)) = sin^{- 1}(sin 5π/3) = 5π/3.

Finally, add multiples of 2π to each of these solutions to get the general form t = 2nπ + 4π/3 and t = 2nπ + 5π/3

where n is integer.

Put t = 5x ⇒ 5x = 2nπ + 4π/3 and 5x = 2nπ + 5π/3

⇒ x = (2nπ + 4π/3)/5 and x = (2nπ + 5π/3)/5, where n ∈ Z.

The general solutions of 3 csc² (5x) = 4 are x = (2nπ + π/3)/5, x = (2nπ + 2π/3)/5, x = (2nπ + 4π/3)/5 and x = (2nπ + 5π/3)/5, where n ∈ Z.