4sin2 x - 3 = 0, find x if 0 ≥ ≥ x ≥ ≥ 90o

Answer Details

The given equation is 4sin²x - 3 = 0. We need to find the value of x such that 0 ≤ x ≤ 90°. To solve this equation, we first need to isolate sin²x by adding 3 to both sides: 4sin²x = 3 Then, we divide both sides by 4: sin²x = 3/4 Now, we take the square root of both sides: sinx = ±√3/2 We have two possible values for sinx: √3/2 and -√3/2. To determine which of these values of sinx is valid for the given range of x, we need to examine the unit circle. The unit circle is a circle with a radius of 1 centered at the origin of the coordinate plane. It is used to visualize the values of sine, cosine, and tangent for all angles in standard position (angles whose vertex is at the origin and whose initial side lies along the positive x-axis). If we draw the unit circle and plot the points corresponding to sinx = √3/2 and sinx = -√3/2, we find that sinx = √3/2 corresponds to an angle of 60° (or π/3 radians) and sinx = -√3/2 corresponds to an angle of 300° (or 5π/3 radians). Since we are looking for a value of x such that 0 ≤ x ≤ 90°, the only valid solution is sinx = √3/2 and x = 60°. Therefore, the answer is x = 60°.