x = nπ + (- 1)n (π/2), where n ∈ Zx = nπ + (- 1)n (5π/6), where n ∈ Zx = nπ + (- 1)n (π/6), where n ∈ Zx = nπ + (- 1)n (π/3), where n ∈ Z

Bạn đang xem: Solve for all values of 2sinx

Concept:

If sin θ = sin α then θ = nπ + (- 1)n α, α ∈ <-π/2, π/2>, n ∈ Z.

 T-Ratios 0° 30° 45° 60° 90° Sin 0 ½ 1/√2 √3/2 1 Cos 1 √3/2 1/√2 1/2 0 Tan 0 1/√3 1 √3 Not defined

Calculation:

Given: 2 sin x – 1 = 0

⇒ sin x = ½ = sin (π/6)

As we know that, if sin θ = sin α then θ = nπ + (- 1)n α, α ∈ <-π/2, π/2>, n ∈ Z.

The general solution of the given equation is: x = nπ + (- 1)n (π/6), where n ∈ Z.
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