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. 60° 0r 120°
- B. 30° 0r 150°
- C. 30° 0r 120°
- D. 60° 0r 150°
Correct Answer: B. 30° 0r 150°
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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