Solve 4cos^2x-3=0 for all real values of x

x=30°, 150°, 210° & 330°

Step-by-step explanation:

4cos²x - 3 = 0

=> Add 3 to both sides, then

4cos²x = 3

=> Divide both sides by 4, then

cos²x = 3/4

=> Square root both sides, then

cos x = ± √ (3/4)

cos x = ± √3/2

cos x = - √3/2 or √3/2

=> Starting cos x = - √3/2, we calculate thus:

x = cos^-1 ( - √3/2)

x = - cos^-1 (√3/2)

=> In trigonometry, cos^-1 of √3/2 is 30° but the - ve arc - cos of a value denotes that the real value of x here can only be obtained from the 2nd & 3rd Quadrant of the unit circle, where the values of cos are - ve, then

2nd Quadrant: 180° - 30° = 150° ...

3rd Quadrant: 180° + 30° = 210° ...

=> Moving on to x = cos^-1 (√3/2), we calculate thus:

x = 30°

=> In the unit circle of the 4 Quadrants, the 4th Quadrant has the cos of all values to be + ve, then

4th Quadrant: 360° - 30° = 330° ...

Hence the real values of x in degrees are 30°, 150°, 210° & 330°

x=30°, 150°, 210° & 330°