Solve for x, 0°≤x≤90° 1. 4 cos^2 2x-3=0 2. 2 sin^2 x - sin x = 0
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1. 4 cos²(2x) - 3 = 0
cos²(2x) = 3/4
cos(2x) = ± √(3/4) = ±√3/2
Remember, cos(Θ) = ±√3/2 when Θ = 30°, 150°
So, 2x = 30° and 2x = 150°
x = 15° and x = 75°
both meet the criterion 0 ≤ x ≤ 90
2. 2 sin²(x) - sin(x) = 0
sin(x) * (2 sin(x) - 1) = 0
So either sin(x) = 0 or 2sin(x) - 1 = 0
sin(x) = 0 when x = 0° or 180°
2sin(x) - 1 = 0 --> sin(x) =1/2
sin(x) = 1/2 when x = 30° or x = 150°
To satisfy the criterion 0 ≤ x ≤ 90,
x = 0° or 30°
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4cos^2(2x)-3 = 0
4cos^2 ^(2x) = 3
(cos2x)^2 = 3/4
Taking square root cos^2x = + or -{(3)^(1/2)}/2 = 30 or-150 deg
Or x = 30/2 =15 or 165 deg
Or, 150/2 =75 or 105 deg
So, x=15 or x=75 are the solutions for which 0<= x<= 90 degree.
(2)
2sin^2x-sinx = 0 oe
six(2sinx-1) = 0.
sinx = 0 when x= 0 or 180 deg
2sinx-1 =0 when sinx = 1/2. when x = 30 deg or x =180+30 =210 degree.
x=0 or x=30 degree are the required solutions in 1st quadrant.
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