Trigonometric Equations

Trigonometric Equations


Solution of trigonometric equations

We know that a trigonometric ratios of a certain angle has one and only one value. But if the value of trigonometric ratio is
given, the angle is not unique. To illustrate the above fact, let us observe the graph of y = sin and find the values of  when
1
y = sin =
2
Y


1

0.8

0.6

0.4
0.2 0.5 0.5

X' X
O 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°
–0.2

–0.4

–0.6

–0.8
–1






Y'
1
From the above graph, when sin = , then the corresponding values of  are 30º, 150º and so on.
2



Rules for finding angles :
1. At first we determine the quadrant where the angle falls. For this we use the CAST table.



(180° –  

S A
(360° + )



T C

(180° + ) (360° – )




Quadrant
I II III IV
Ratios
Sin and cosec +ve +ve –ve –ve

Cos ans sec +ve –ve –ve +ve
Tan and cot +ve –ve +ve –ve
2. Find the least positive angle of the trigonometric function in the first quadrant for the given relation.
For example
1
(i) For sin = , the least positive angle in the first quadrant is 45º
2
1 1
(ii) For cos = – , the least positive angle in the first quadrant for cos = is 60º
2
2
1 1
(iii) For tan = – , the least positive angle in the first quadrant for tan = is 30º
3 3
(iv) For sin = –1, the least positive angle in the first quadrant for sin = 1 is 90º

3. When  is the least positive angle in the first quadrant then the angle in the second quadrant = (180º – )
The angle in the third quadrant = (180º + )
The angle in the fourth quadrant = (360º – )

To find the angle more them 360º we use (360º + ) and so on.

Note: To find the angle in the second, third and fourth quadrant, we add or subtract the least positive angles of first
quadrant in even multiple of 90º.
Worked out Examples

1. Solve (0º    360º)
(a) 2 sin = 1 (b) 2cos + 3 = 0 (c) 3 tan – 3 = 0
Solution
(a) Here, 2 sin = 1
1
nd
st
or, sin = (Here, sin is positive. So,  lies in 1 and 2 quadrants)
2
or, sin = sin45º, sin(180º – 45º)
  = 45º, 135º

(b) Here, 2cos + 3 = 0
– 3
nd
rd
or, cos = (Here cos is negative. So  lies in 2 and 3 quadrant)
2
or, cos = – cos30º
or, cos = cos(180º – 30º), cos (180º + 30º)
  = 150º, 210º

(c) Here, 3 tan – 3 = 0
st
rd
or, tan = 3 (Here, tan is positive. So  lies in 1 and 3 quadrants)
or, tan = tan60º, tan (180º + 60º)
  = 60º, 240º
2. Solve (0º    180º)
2
(a) 4sin  – 3 = 0 (b) 9 tan  – 3 = 0
2
Solution:
2
(a) Here, 4sin  – 3 = 0
3
or, sin  =
2
4
3
or, sin = ±
2

Taking +ve sign,
3
sin =
2
or, sin = sin60º, sin (180 – 60º)
  = 60º, 120º

Taking –ve sign

– 3
sin = 2
or, sin = – sin60°
or, sin = sin(180º + 60º), sin (360º – 60º)
  = 240º, 300º
But 0º    180º
Hence,  = 60º, 120º

2
(b) 9 tan  – 3 = 0
1
2
or, tan  =
3
1
or, tan = ±
3

Taking +ve sign,
1
tan =
3
or, tan = tan30º, tan (180 + 30º)
  = 30º, 210º

Taking –ve sign
–1
tan =
3
or, tan = – tan30º
or, tan = tan (180º – 30º), tan (360º – 30º)
  = 150º, 330º
But, 0º    180º
Hence,  = 30º, 150º

3. Solve : (0º    180º)
(a) sin2 + cos = 0 (b) tan = cot5
Solution :
(a) Here, Sin2 + cos = 0
or, 2sin.cos + cos = 0
or, cos (2sin + 1) = 0

Either, cos = 0
or, cos = cos90º
  = 90º
OR, 2sin + 1 = 0
–1
or, sin =
2
or, sin = – sin30º
or, sin = sin (180º + 30º), sin (360º – 30º)
  = 210º, 330º
But 0º    180º
  = 90º

(b) Here, tan = cot5
or, tan = tan (90º – 5)
or,  = 90º – 5
or, 6 = 90º
  = 15º

4. Solve (0º    360º)
(a) (3sin – 4) (2sin + 3 ) = 0
(b) (2sec + 3) (tan + 3 ) = 0
Solution :
(a) Here, (3sin – 4) (2sin  + 3 ) = 0
Either, 3sin – 4 = 0
4
or, sin = (Rejected) [since value of sin and cos lies from – 1 to +1]
3
OR 2sin + 3 = 0
– 3
or, sin = 2

or, sin = – sin60º
or, sin = sin(180º + 60º), sin(360º – 60º)
Hence,  = 240º, 300º

(b) (2sec + 3) (tan + 3 ) = 0
Solution :
Here, (2sec + 3) (tan + 3 ) = 0
Either, 2sec + 3 = 0
–3
or, sec =
2
–3
–1  
  = sec
 
2

OR, tan + 3 = 0
or, tan = – 3
or tan = – tan60º
or, tan = tan (180º – 60º), tan (360º – 60º)
  = 120º, 300º
–3
–1  
Hence,  = sec   , 120º, 300º
2

5. Solve (0º    360º)
(a) 2sin  + 3cos = 3 (b) tan  – 3sec + 3 = 0
2
2
(c) sec.tan = 2 (d) 3 sin + cos = 2
(e) cos – 3 sin = 2 (f) sin2 + sin4 = cos + cos3

Solution :
2
(a) 2sin  + 3cos = 3
2
Here, 2sin  + 3cos = 3
or, 2(1 – cos ) + 3cos – 3 = 0
2
or, 2 – 2cos  + 3cos – 3 = 0
2
or, 2cos  – 3cos +1 = 0
2
or, 2cos  – 2cos – cos + 1 = 0
2
or, 2cos (cos – 1) – 1 (cos – 1) = 0
or, (cos – 1) (2cos – 1) = 0

Either, cos – 1 = 0
or, cos = 1
or, cos = cos0º, cos (360º – 0º)
  = 0º, 360º
OR, 2cos – 1 = 0
1
or, cos =
2
or, cos = cos60º, cos (360º – 60º)
  = 60º, 300º

Hence,  = 0º, 60º, 300º, 360º

2
(b) tan  – 3sec + 3 = 0
or, sec  – 1 – 3sec + 3 = 0
2
2
or, sec  – 3sec + 2 = 0
2
or, sec  – 2sec – sec + 2= 0
or, sec (sec – 2) –1 (sec –2) = 0
or (sec – 1) (sec – 2) = 0

Either, sec – 1 = 0
or, sec = 1
or, sec = sec0º, sec (360º – 0º)
  = 0º, 360º
OR, sec – 2 = 0
or, sec = 2
or, sec = sec 60º, sec (360º – 60º)
  = 60º, 300º
Hence,  = 0º, 60º, 300º, 360º

(c) sec.tan = 2
Here, sec.tan = 2
1 sin
or, . = 2
cos cos
or, sin = 2 (1 – sin )
2
or, 2 sin  + sin – 2 = 0
2
2
or, 2 sin  + 2sin – sin – 2 = 0
or 2 sin (sin + 2 ) – 1 (sin + 2 ) = 0
or (sin + 2 ) ( 2 sin – 1) = 0

Either, sin + 2 = 0
or, sin = 2 (rejected)
OR, 2 sin – 1 = 0
1
or, sin =
2
or, sin = sin45º, sin(180º – 45º)
  = 45º, 135º
Hence,  = 45º, 135º

(d) 3 sin + cos = 2 … (i)
2
2
Here, (coefficient of sin) + (coefficient of cos)
2
= ( 3) + (1)
2
= 2

Now, dividing equation (i) on both sides by 2, we get
3 1 2
sin . + cos . =
2 2 2
1
or, sin.cos30º + cos.sin30º =
2
or, sin ( + 30º) = sin45º, sin(180 – 45)
or,  + 30º = 45º, 135º
  = 15º, 105º
Hence,  = 15º, 105º

(e) cos – 3 sin = 2
or, 3 sin – cos – 2 … (i)
2
Here, (coefficient of sin) + (coefficient of cos) = ( 3) + (1)
2
2
2
= 3 + 1
= 2

Now, dividing both sides of equation (i) by 2, we get
3 cos –2
sin – =
2 2 2
or, sin.cos30º – cos.sin30º = –1
or, sin ( – 30º) = – sin90º
or, sin ( – 30º) = sin (360º – 90º)
or,  – 30º = 270º
  = 300º
Hence,  = 300º
(f) sin2 + sin4 = cos + cos3
Here, sin2 + sin4 = cos + cos3
2 + 4 4 – 2 3 +  3 – 
or, 2sin . cos = 2cos . cos
2 2 2 2
or, 2sin 3.cos = 2cos2.cos
or, 2sin3.cos – 2cos2.cos = 0
or, 2cos (sin3 – cos2) = 0

Either, 2cos = 0
or, cos = 0
or, cos = cos90º, cos (360º – 90º)
  = 90º, 270º

OR, sin3 – cos2 = 0
or, sin3 = cos2
or, sin3 = sin (90°– 2)
or, 3 = 90° – 2
or, 5 = 90º
  = 18º
Hence,  = 18º, 90º, 270º

SEE Questions
Short Qustions
2
1. (a) olb 2cos  = – 3 cos eP 0° b]lv 180° leq kg]{  sf] dfg lgsfNg'xf];\ .
2
If 2cos  = – 3 cos find the values of  which lies between 0° to 180°. 2059 R
  1
2
(b) olb cos – cos + = 0 eP 180° leq kg]{  sf] dfg lgsfNg'xf];\ .
2 2 4
  1
2
If cos – cos + = 0, find the value of  which lies between 0° and 180°. 2060 R
2 2 4
2. xn ug'{xf];\ (Solve): {0°  (x)  180°}
(a) sin  = cos  2063 R (f) cos 2x = sin x 2064 R-II
2
(b) 2 cos  – 1 = 0 2063 Supp. (g) 3 tan  = 3 2063 R-II
2
(c) 2 cos  = 1 Mode 2063 (h) cot x = 3 2065 R
(d) 4 sin  = 3 cosec  2064 R-II (i) sin – tan = 0 2062 K
2
(e) 2 cos  – 3cos  = 0 2064 Supp. (j) tan  = cot  2064 Supp.


Additional Questions


1. xn ug'{xf];\ (Solve): {0  (/A/ 360}
(a) 2sin  3 = 0 (b) 3 tan + 1 = 0 (c) 2secA  2 = 0
2
(d) 2cos   1 = 0 (e) sin   0.75 = 0 (f) 3tan   3 = 0
2
2
2. xn ug'{xf];\ (Solve): (0    90°)
(a) sin3 = cos7. (b) tan = cot. (c) tan + cot = 2.
(d) tan = 2sin. (e) sin2  cos = 0

3. xn ug'{xf];\ (Solve): (0°    180°)
(a) (2sin  1) cos = 0 (b)(sin + 1) (2sin  1) = 0

(c) sin  2sin . cos = 0 (d) 2sin . cos  sin = 0
2
  3
2
(e) cos + cos  = 0
2 2 4
4. xn ug'{xf];\ (Solve): (0°  A  180°)
1 3
o
(a) sinA . cos30  cosA . sin30 = (b) sin45 . cosA + cos45 . sinA =
2 2
1 1
o
(c) cos60 . cosA  sin60 . sinA = (d) cosA . cos45 + sinA . sin45 = .
2 2
SEE Questions
1. xn ug'{xf];\ (Solve):
2
(a) 2 cos  = 3 sin  (0°    180°) 2057 Supp.
1

2
(b) cos  – sin  = (0°   180°) 2061 R
4
2
(c) 2 cos  + sin  = 2 (0°    180°) 2061 Supp.
2
(d) 1 + cos  = 2 sin  (0°    180°) 2058 Supp.
2
2
(e) cos x = 3 sin x + 4 cos x, (0°  x  360°) 2062 R
2
(f) 3 cos 2x – 1 = 2 sin x (0°  x  180°) 2064R-II
2
2
(g) 7 sin  + 3 cos  = 4 (0°    360°) 2065 E
2. xn ug'{xf];\ (Solve):
(a) cos 2 = sin  (0°   360°) 2058 R

(b) sec . tan  = 2 (0°    180°) 2060 R
3. xn ug'{xf];\ (Solve):

(a) 3 sin  – cos  = 2 (0°   180°) 2057 R

(b) sin 3x + sin x = 2 sin x (0°  x  180°) 2063 R-II
(c) cos 3x + cos x = 2 cos x (0°  x  180°) Model 2063
(d) cos 3 + cos  = cos 2 (0°  x  180°) 2064 Supp.
(e) sin 4x + sin 2x = cos x (0°  x  360°) 2063 Supp.
4. xn ug'{xf];\ (Solve): sin 3 + sin  = sin 2 (0°   180°) 2061 R

3
5. xn ug'{xf];\ (Solve): 3 cot A = – 1 (0°  A  360°) 2064 R
sin A

 1 
2
6. xn ug'{xf];\ (Solve): cot x +  3 +  cot x = – 1 (0°  x  360°) 2065 R-II
  3

Answers
1. (a) 30° or 150° (f) 30°, 150° 3. (a) 75°, 165° 4. 0°, 60°, 90°, 180°
(b) 30°, 150° (g) 30°,150°,210°,330° (b) 0°, 180°, 360°
(c) 0°, 180° (c) 0°, 90°, 180° 5. 0°, 60°, 360°
(d) 60°, 180° 2. (a) 30° (d) 45°, 60°, 135° 6. 120°, 150°, 300°, 330°
(e) 120°, 240° (b) 45°, 135° (e) 10°, 150°, 130°