5.6c Solving Trigonometric Equation (Form Quadratic Equation in sinx/ cosx/ tanx/ cosecx/ secx/ cotx)

(C) Solving Trigonometric Equation (Form Quadratic Equation in sinx/ cosx/ tanx/ cosecx/ secx/ cotx)

Example:
Find all the angles between 0° and 360° that satisfy each of the following equations.
(a) 3 sin² x – 2 sin x – 1 = 0
(b)  2 sin x = cosec x + 1
(c) 5 sin² x = 2 (1 + cos x)
(d) 2 sec x = 1 + cos x
(e)  2 cot² x + 8 = 7 cosec x

Solution:
(a) 3 sin² x – 2 sin x – 1 = 0
(3 sin x + 1)(sin x – 1) = 0
sin x = –, sin x = 1
sin x = –
basic angle = 19.47°
x = 180° + 19.47°, 360° – 19.47°
x = 199.47°, 340.53
sin x = 1, x = 90°
Hence x = 90°, 199.47°, 340.53°

(b) 2 sin x = cosec x + 1
2sinx= 1 sinx +1
2 sin ² x = 1 + sin x
2 sin ² x – sin x – 1 =0
(2 sin x + 1)(sin x – 1) = 0
sin x = –½, sin x = 1
sin x = –½
basic angle = 30°
x = 180° + 30°, 360° – 30°
x = 210°, 330°
sin x = 1, x = 90°
Hence x = 90°, 210°, 330°

(c) 5 sin² x = 2 (1 + cos x)
5 (1 – cos² x) = 2 + 2 cos x
5 – 5 cos² x – 2 – 2 cos x = 0
– 5 cos² x – 2 cos x + 3 = 0
5 cos² x + 2 cos x – 3 = 0
(5 cos x – 3)(cos x + 1) = 0
cosx= 3 5 , cosx=1 cosx= 3 5
basic angle = 53.13°
x = 53.13°, 360° – 53.13°
x = 53.13°, 306.87°
cos x = – 1
x = 180°
Hence x = 53.13°, 180°, 306.87°

(d) 2 sec x = 1 + cos x
2 cosx =1+cosx
2 = cos x + cos² x
cos² x + cos x – 2 = 0
(cos x – 1)(cos x + 2) = 0
cos x = 1
x = 0°, 360°
cos x = –2 (not accepted)
Hence x = 0°, 360°

(e) 2 cot² x + 8 = 7 cosec x
2 (cosec² x – 1) + 8 = 7 cosec x
2 cosec² x – 2 – 7 cosec x + 8 = 0
2 cosec2 x – 7 cosec x + 6 = 0
(2 cosec x – 3)(cosec x – 2) = 0
cosecx= 3 2 ,   cosecx=2 sinx= 2 3 ,       sinx= 1 2 sinx= 2 3 basic angle =  41.81 x= 41.81 , 180 41.81 sinx= 1 2 basic angle = 30 x= 30 , 180 30 Hence x =  30 ,  41.81 ,  138.19 ,  150