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A trig equation (quadratic in sin)

Trigonometry
A-level Maths (9MA0)
ORIGINAL

Spot the hidden quadratic in sin x, solve it, then find every angle in the range.

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Question

1.

Solve 2sin⁡2x+sin⁡x−1=02\sin^2 x + \sin x - 1 = 0 for 0≤x<360∘0 \le x < 360^\circ.

Every step worked, with the reasoning.

  1. 1
    Let s=sin⁡xs = \sin x: 2s2+s−1=02s^2 + s - 1 = 0.

    It's a quadratic in sin⁡x\sin x.

  2. 2
    (2s−1)(s+1)=0(2s - 1)(s + 1) = 0

    Factorise the quadratic.

  3. 3
    sin⁡x=12\sin x = \dfrac{1}{2} or sin⁡x=−1\sin x = -1

    Solve each factor for sin⁡x\sin x.

  4. 4
    sin⁡x=12:  x=30∘, 150∘\sin x = \tfrac{1}{2}:\ \ x = 30^\circ,\ 150^\circ

    Principal value 30∘30^\circ, and 180∘−30∘=150∘180^\circ - 30^\circ = 150^\circ, both in range.

  5. 5
    sin⁡x=−1:  x=270∘\sin x = -1:\ \ x = 270^\circ

    The only angle in range with sin⁡x=−1\sin x = -1.

Answer: x=30∘, 150∘, 270∘x = 30^\circ,\ 150^\circ,\ 270^\circ

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