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:

Concept:

Isolate

\sin^2\theta

, then take the square root and use the given range of

\theta

to find all valid solutions.

Explanation:

Start with the equation:

4\sin^2\theta + 1 = 2

Subtract 1 from both sides:

4\sin^2\theta = 1

Divide both sides by 4:

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

Take the square root of both sides:

\sin\theta = \pm \frac{1}{2}

Since

0^\circ < \theta < 180^\circ

\sin\theta

is positive, so:

\sin\theta = \frac{1}{2}

Thus:

\theta = \sin^{-1}\left(\frac{1}{2}\right)

The angles between

0^\circ

and

180^\circ

with

\sin\theta = \frac{1}{2}

are:

\theta = 30^\circ

\theta = 150^\circ

Answer:

Option B:

30^\circ

150^\circ

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