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°

goog may have it for you

damm bro the struggle

step-by-step explanation: