Solve for x, 0°≤x≤90° 1. 4 cos^2 2x-3=0 2. 2 sin^2 x - sin x = 0

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°

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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Solve for x, 0°≤x≤90° 1. 4 cos^2 2x-3=0 2. 2 sin^2 x - sin x = 0

Expert Answers

kjcdb8er

| Certified Educator

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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Student Comments

neela | Student

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.