Question 4:
Solve the equation 3 sin A cos A – cos A = 0 for 0° ≤ A ≤ 360°.
Solution:
3 sin A cos A – cos A = 0
cos A (3 sin A – 1) = 0
cos A = 0 or sin A = ⅓
cos A = 0
A = 90°, 270°
sin A = ⅓
Basic angle = 19°28'
A = 19°28', 180° – 19°28'
A = 19°28', 160°32'
Hence A = 19°28', 90°, 160°32', 270°.
Solve the equation 3 sin A cos A – cos A = 0 for 0° ≤ A ≤ 360°.
Solution:
3 sin A cos A – cos A = 0
cos A (3 sin A – 1) = 0
cos A = 0 or sin A = ⅓
cos A = 0
A = 90°, 270°
sin A = ⅓
Basic angle = 19°28'
A = 19°28', 180° – 19°28'
A = 19°28', 160°32'
Hence A = 19°28', 90°, 160°32', 270°.
Question 5:
Solve the equation 4 sin (x – π) cos (x – π) = 1 for 0o ≤ x ≤ 360o.
Solution:
4 sin (x – π) cos (x – π) = 1
2 [2 sin (x – π) cos (x – π)] = 1
2 sin (x – π) cos (x – π) = ½
sin 2(x – π) = ½ ← (sin 2x= 2 sinx cosx)
sin 2(x – 180o) = ½ ← (π rad = 180o)
sin (2x – 360o) = ½
sin 2x cos 360o – cos 2x sin 360o = ½
sin 2x (1) – cos 2x (0) = ½ ← (cos 360o = 1, sin 360o = 0)
sin 2x = ½
basic angle = 30o ← (special angle, sin 30o= ½)
2x = 30o, 150o, 390o, 510o
x = 15o, 75o, 195o, 255o