solve!!!!!!!!!!!!!!!!!!!!!

Question

2sin^2 x - 3sin x + 1 = 0

Answer 1

The trigonometric equation is 2sin² x - 3sin x + 1 = 0.

By factoring by grouping.

2sin² x - 2sin x - sin x + 1 = 0

2sin x(sin x - 1) - 1(sin x - 1) = 0

Factor : (sin x - 1)(2sin x - 1) = 0

Apply zero product property.

sin x - 1 = 0  and  2sin x - 1 = 0

sin x = 1 and  sin x = 1/2.

• sin (x) = 1.

sin(x) = sin(π/2)

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

⇒ x = nπ + (- 1)ⁿπ/2.

• sin (x) = 1/2.

sin(x) = sin(π/6)

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

⇒ x = nπ + (- 1)ⁿπ/6.

The solutions of the given equation are x = nπ + (- 1)ⁿπ/2 and x = nπ + (- 1)ⁿπ/6, where n is an integer.