Question on: JAMB Mathematics - 2019
If \(4\sin^2 x - 3 = 0\), find the value of x, when 0° \(\leq\) x \(\leq\) 90°
A
90°
B
45°
C
60°
D
30°
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Correct Option: C
\(4\sin^2 x - 3 = 0\)
\(4 \sin^2 x = 3 \implies \sin^2 x = \frac{3}{4}\)
\(\sin x = \frac{\sqrt{3}}{2}\)
\(\therefore x = \sin^{-1} (\frac{\sqrt{3}}{2})\)
x = 60°
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