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

2 Answers | Add Yours

kjcdb8er's profile pic

kjcdb8er | Teacher | (Level 1) Associate Educator

Posted on

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°

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

 

We’ve answered 318,994 questions. We can answer yours, too.

Ask a question