Trigonometric Functions Short Questions (Question 1 - 4)


Question 1:
Solve the equation 3 cos 2A = 8 sin A – 5 for 0°  A ≤ 360°.

Solution:
3 cos 2A = 8 sin A – 5
3(1–2 sin2 A) = 8 sin A – 5
3 – 6 sin2 A = 8 sin A – 5
6 sin2 A + 8 sin A – 8 = 0
3 sin2 A + 4 sin A – 4 = 0
(3 sin A – 2)(sin A + 2) = 0  ( Factorise the equation )
3 sin A – 2 = 0
sin A 2 3 ( sinA is positive in the first and second quadrants. )
A41°49’, 138°11’
Or
sin A + 2 = 0
sin A = –2 (no solution)

Hence A41°49’, 138°11’.


Question 2:
Solve the equation 2 cos 2x – cos x – 1 = 0 for 0°  x ≤ 360°.

Solution:
2 cos 2x – cos x – 1 = 0
2 (2 cos2 x – 1) – cos x – 1 = 0
4 cos2 x – 2 – cos x – 1 = 0
4 cos2 x – cos x – 3 = 0
(4 cos x + 3)(cos x – 1) = 0
cos x =  3 4
basic angle = 41°24’
x = 138°36’, 221°24’
or
cos x = 1,
x = 0°, 360°

Hence x0°, 138°36’, 221°24’, 360°




Question 3:
Solve the equation 6 sec² x – 13 tan x = 0, 0°   x  360°.

Solution:
6 sec² x – 13 tan = 0
6 (1 + tan²x) – 13 tan x = 0
6 tan²x – 13 tan x + 6 = 0
(3 tan x – 2)(2 tan x – 3) = 0
tan x = 2/3   or   tan x = 3/2

tan x = 2/3
Basic angle = 33.69°
x = 33.69°, 180° + 33.69°
x = 33.69°, 213.69° 
Or
tan x = 3/2
Basic angle = 56.31°
x = 56.31°, 180° + 56.31°
x = 56.31°, 236.31°

Hence x = 33.69°, 56.31°, 213.69°, 236.31°. 




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

cos A = 0
A = 90°, 270°

sin A
Basic angle = 19°28'
A = 19°28', 180° – 19°28'
A = 19°28', 160°32'

Hence A = 19°28', 90°, 160°32', 270°.