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Solve: 4sin\(^2\)θ + 1 = 2, where 0º < θ < 180º

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Further Mathematics WAEC 2022

Solve: 4sin\(^2\)θ + 1 = 2, where 0º < θ < 180º

  • 60º 0r 120º
  • 30º 0r 150º checkmark
  • 30º 0r 120º
  • 60º 0r 150º

The correct answer is: B

Explanation

4sin\(^2\)θ + 1 = 2

4sin\(^2\)θ  = 2 - 1

4sin\(^2\)θ = 1

\(\sqrt sin^2θ\) = \(\sqrt \frac{1}{4}\)

sinθ = \(\frac{1}{2}\)

θ = \(sin^{-1} \frac{1}{2}\)

θ = 30º 0r 150º

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