If 4sin2x−3=0 4 sin 2 ⁡ x − 3 = 0 , find the value of x, when 0° ≤ ≤ x ≤ ≤ 90°

If $4{\sin }^{2}x-3=0$, find the value of x, when 0° $\le$ x $\le$ 90°

Answer Details

We start by solving the equation: 4sin2x - 3 = 0 Adding 3 to both sides, we get: 4sin2x = 3 Dividing both sides by 4, we get: sin2x = 3/4 Using the identity sin2x = 2sinx cosx, we can write: 2sinx cosx = 3/4 Now, we need to find values of x for which this equation holds true. Since sinx and cosx are both positive or both negative in the first and fourth quadrants, we can restrict our search to these quadrants. Let's consider the first quadrant. Here, sinx and cosx are both positive. We can rewrite our equation as: 2sinx cosx = 3/4 sinx cosx = 3/8 We can now use the fact that sin(90 - x) = cosx to get: sin(90 - x) sinx = 3/8 cosx sinx = 3/8 Using the identity 2sinx cosx = sin2x, we get: sin2x/2 = 3/8 sin2x = 3/4 Solving for x, we get: 2x = sin^-1(3/4) x = sin^-1(3/4)/2 Using a calculator, we get: x ≈ 48.59° Since sinx is an odd function, we can also get solutions in the fourth quadrant by taking the negative of our solution in the first quadrant: x ≈ -48.59° + 360° = 311.41° However, we are only interested in values of x between 0° and 90°, so the only solution in this range is: x ≈ 48.59° Therefore, the answer is option (C) 60° (rounded to the nearest degree).