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

Further Mathematics
WAEC 2022

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

  • A. 60º 0r 120º
  • B. 30º 0r 150º
  • C. 30º 0r 120º
  • D. 60º 0r 150º
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Correct Answer: Option 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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