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Sonny Rhodes
Mathematics
17 March, 01:34
Solve 4cos^2x-3=0 for all real values of x
+5
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Juan Wilkins
17 March, 01:41
0
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°
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Castaneda
17 March, 01:58
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x=30°, 150°, 210° & 330°
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