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JAMB Mathematics 2019 Paper

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Question : 18

Total: 73

If

4\sin^2 x - 3 = 0

, find the value of x, when 0°

\leq

\leq

90°

90°
45°
60°
30°

Solution:

Concept:

Solve the trigonometric equation for

x

within the given range

0^\circ \leq x \leq 90^\circ

.

Explanation:

Start with

4\sin^2 x - 3 = 0

.

Add

3

to both sides:

4\sin^2 x = 3

.

Divide both sides by

4

:

\sin^2 x = \frac{3}{4}

.

Take the positive square root because sine is positive in the first quadrant:

\sin x = \frac{\sqrt{3}}{2}

.

So,

x = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)

.

Since

\sin 60^\circ = \frac{\sqrt{3}}{2}

, we get

x = 60^\circ

.

Answer:

x = 60^\circ

(Option C).

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