How do I find all solution of the equation

Question

sin^2 (3x) = 1 ?

Answer 1

The trigonometric equation is sin² (3x) = 1.

⇒ sin (3x) = ± 1

sin (3x) = - 1 and sin (3x) = + 1.

• sin (3x) = - 1.

sin(3x) = sin(3π/2)

The genaral solution of sin(θ) = sin(α) is θ = nπ + (- 1)ⁿα, where n is an integer.

 3x = nπ + (- 1)ⁿ(3π/2)

⇒ x = [nπ + (- 1)ⁿ(3π/2)]/3.

• sin (3x) = 1.

sin(3x) = sin(π/2)

The genaral solution of sin(θ) = sin(α) is θ = nπ + (- 1)ⁿα, where n is an integer.

3x = nπ + (- 1)ⁿ(π/2)

⇒ x = [nπ + (- 1)ⁿ(π/2)]/3.

The solutions of the given equation are x = [nπ + (- 1)ⁿ(3π/2)]/3 and x = [nπ + (- 1)ⁿ(π/2)]/3, where n is an integer.