Solve the equation sin 5/2ø= 1/2 for 0⁰<ø< 180⁰

Mathematics

ScanSolve AI Answer

Answer

12∘,60∘,156∘12^\circ, 60^\circ, 156^\circ

Step-by-step solution

4 steps
  1. The general solution for sin⁡ϕ=12\sin \phi = \frac{1}{2} is

    ϕ=30∘+360∘norϕ=150∘+360∘n,\phi = 30^\circ + 360^\circ n \quad \text{or} \quad \phi = 150^\circ + 360^\circ n,

    where nn is any integer.

  2. Substitute ϕ=52θ\phi = \frac{5}{2} \theta.

    52θ=30∘+360∘nor52θ=150∘+360∘n.\frac{5}{2} \theta = 30^\circ + 360^\circ n \quad \text{or} \quad \frac{5}{2} \theta = 150^\circ + 360^\circ n.
  3. Solve for θ\theta.

    First equation:

    θ=25(30∘+360∘n)=12∘+144∘n.\theta = \frac{2}{5} (30^\circ + 360^\circ n) = 12^\circ + 144^\circ n.

    Second equation:

    θ=25(150∘+360∘n)=60∘+144∘n.\theta = \frac{2}{5} (150^\circ + 360^\circ n) = 60^\circ + 144^\circ n.
  4. Apply 0∘<θ<180∘0^\circ < \theta < 180^\circ.

    For θ=12∘+144∘n\theta = 12^\circ + 144^\circ n:

    • n=0n=0: θ=12∘\theta = 12^\circ
    • n=1n=1: θ=156∘\theta = 156^\circ
    • n=2n=2: θ=300∘>180∘\theta = 300^\circ > 180^\circ
    • n=−1n=-1: θ=−132∘<0∘\theta = -132^\circ < 0^\circ

    For θ=60∘+144∘n\theta = 60^\circ + 144^\circ n:

    • n=0n=0: θ=60∘\theta = 60^\circ
    • n=1n=1: θ=204∘>180∘\theta = 204^\circ > 180^\circ
    • n=−1n=-1: θ=−84∘<0∘\theta = -84^\circ < 0^\circ

    Solutions: θ=12∘,60∘,156∘\theta = 12^\circ, 60^\circ, 156^\circ

    12∘,60∘,156∘12^\circ, 60^\circ, 156^\circ
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