If \(4\sin^2 x - 3 = 0\), find the value of x, when 0° \(\leq\) x \(\leq\) 90°
The correct answer is C. 60°
To find the value of x, we first solve the equation \(4\sin^2 x - 3 = 0\) for sin^2x:
\(4\sin^2 x = 3\).
Now, solve for sin x:
\(\sin^2 x = \frac{3}{4}\).
Take the square root of both sides:
\(\sin x = \sqrt{\frac{3}{4}}\).
\(\sin x = \frac{\sqrt{3}}{2}\).
Now, find the angle whose sine is \(\frac{\sqrt{3}}{2}\). This is a standard angle, and we know that \(\sin 60° = \frac{\sqrt{3}}{2}\).
So, \(x = 60°\)
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