Solve: 4sin²θ + 1 = 2, where 0º

Correct Answer is : 30º 0r 150º

Correct Answer is : 30º 0r 150º

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WAEC Further Mathematics 2022 Paper

Question : 35

Total: 45

Solve: 4sin

^{2}

θ + 1 = 2, where 0º < θ < 180º

Solution:

4sin $^{2}$ θ + 1 = 2

4sin $^{2}$ θ = 2 - 1

4sin $^{2}$ θ = 1

$\sqrt{\sin^2\theta}$ = $\sqrt{\frac{1}{4}}$

sinθ = $\frac{1}{2}$

θ = $\sin^{-1}\frac{1}{2}$

θ = 30º 0r 150º

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