Subject : Mathematics Topic : TRIGONOMETRI EQUATIONS
Page : 1 of 15 TRIG. EQUATIONS
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STUDY PACKAGE
Subject : Mathematics Topic : TRIGONOMETRI EQUATIONS
R
TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: 0 903 903 7779, 98930 58881 , BHOPAL
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Index 1. Theory 2. Short Revision 3. Exercise (Ex. 1 + 5 = 6) 4. Assertion & Reason 5. Que. from Compt. Exams 6. 39 Yrs. Que. from IIT-JEE(Advanced) 7. 15 Yrs. Que. from AIEEE (JEE Main)
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Page : 2 of 15 TRIG. EQUATIONS
Trigonometric Equation
1 . Trigonometric Equation :
An equation involving one or more trigonometric ratios of an unknown angle is called a trigonometric
equation.
2 . Solution of Trigonometric Equation :
A solution of trigonometric equation is the value of the unknown angle that satisfies the equation.
e.g. if sin = 1 2
3 9 11 = 4 , 4 , 4 , 4 , ...........
Thus, the trigonometric equation may have infinite number of solutions (because of their periodic nature) and
can be classified as :
(i)
Principal solution
(ii) General solution.
2.1 Principal solutions:
The solutions of a trigonometric equation which lie in the interval
[0, 2) are called Principal solutions.
1 e.g Find the Principal solutions of the equation sinx = 2 . Solution.
TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: 0 903 903 7779, 98930 58881 , BHOPAL
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1
sinx =
2
there exists two values
5
1
i.e.
6 and 6 which lie in [0, 2) and whose sine is 2
1
Principal solutions of the equation sinx = 2 are 6 ,
2.2 General Solution :
5 6 Ans.
The expression involving an integer 'n' which gives all solutions of a trigonometric equation is called General solution.
General solution of some standard trigonometric equations are given below.
3 . General Solution of Some Standard Trigonometric Equations :
(i) If sin = sin
(ii) If cos = cos
(iii) If tan = tan
(iv) If sin? = sin?
(v) If cos? = cos?
(vi) If tan? = tan?
Some Important deductions :
(i) sin = 0
(ii) sin = 1
(iii) sin = ? 1
(iv) cos = 0
(v) cos = 1
(vi) cos = ? 1
(vii) tan = 0
= n + (-1)n = 2n ?
= n+ = n ? , n . = n ? , n . = n ? , n .
where
-
2
,
2
,
where [0, ],
where - , , 2 2
n . n . n .
[ Note: is called the principal angle ]
= n,
n
= (4n + 1) 2 , n
= (4n ? 1) , n
2
= (2n + 1) , n
2
= 2n,
n
= (2n + 1), n
= n,
n
Solved Example # 1 Solve sin = 3 . 2
Solution.
Page : 3 of 15 TRIG. EQUATIONS
sin = 3
2
sin = sin 3
= n + (? 1)n 3 , n
Solved Example # 2
2 Solve sec 2 = ? 3 Solution.
2
sec 2 = ?
3
cos2 = ? 3
2
2 = 2n ? 5 , n
6
5 = n ? 12 , n
Ans.
5 cos2 = cos 6 Ans.
TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: 0 903 903 7779, 98930 58881 , BHOPAL
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Solved Example # 3
Solve tan = 2
Solution.
Let
tan = 2
............(i)
2 = tan
tan = tan
= n + , where = tan?1(2), n
Self Practice Problems:
1.
Solve cot = ? 1
1
2.
Solve cos3 = ? 2
2n 2
Ans. (1)
= n ? 4 , n
(2)
3 ? 9 ,n
Solved Example # 4
1 Solve cos2 = 2 Solution.
1
cos2 =
2
cos2 =
1 2
2
cos2 = cos2
4
= n ? 4 , n Ans.
Solved Example # 5
Solve 4 tan2 = 3sec2
Solution.
4 tan2 = 3sec2
.............(i)
For equation (i) to be defined (2n + 1) 2 , n
equation (i) can be written as:
4 sin2
3
cos2 = cos2
4 sin2 = 3
(2n + 1) 2 , n
cos2 0
sin2 =
3 2 2
Page : 4 of 15 TRIG. EQUATIONS
sin2 = sin2 3
= n ? 3 , n Ans.
Self Practice Problems :
1.
Solve 7cos2 + 3 sin2 = 4.
2.
Solve 2 sin2x + sin22x = 2
Ans. (1)
n ? 3 , n
(2) (2n + 1) , n
or
n ? , n
2
4
Types of Trigonometric Equations :
TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: 0 903 903 7779, 98930 58881 , BHOPAL
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Type -1
Trigonometric equations which can be solved by use of factorization.
Solved Example # 6
Solve (2sinx ? cosx) (1 + cosx) = sin2x.
Solution.
(2sinx ? cosx) (1 + cosx) = sin2x
(2sinx ? cosx) (1 + cosx) ? sin2x = 0
(2sinx ? cosx) (1 + cosx) ? (1 ? cosx) (1 + cosx) = 0
(1 + cosx) (2sinx ? 1) = 0
1 + cosx = 0
or
2sinx ? 1 = 0
cosx = ? 1
1
or
sinx =
2
x = (2n + 1), n or
sin x = sin 6
Solution of given equation is
(2n + 1), n Self Practice Problems :
or
n + (?1)n , n
6
x = n + (? 1)n
6 , n
Ans.
x
1.
Solve cos3x + cos2x ? 4cos2 = 0
2
2.
Solve cot2 + 3cosec + 3 = 0
Ans. (1) (2)
Type - 2
(2n + 1), n 2n ? , n or
2
n + (?1)n + 1 , n 6
Trigonometric equations which can be solved by reducing them in quadratic equations.
Solved Example # 7
Solve 2 cos2x + 4cosx = 3sin2x
Solution.
2cos2x + 4cosx ? 3sin2x = 0 2cos2x + 4cosx ? 3(1? cos2x) = 0 5cos2x + 4cosx ? 3 = 0
cos
x
-
-
2
+ 5
19
cos
x
-
-
2
- 5
19
= 0
cosx [? 1, 1] x R
........(ii)
- 2 - 19
cosx
5
equation (ii) will be true if
- 2 + 19
cosx =
5
- 2 + 19
cosx = cos, where cos =
5
x = 2n ? where
= cos?1
-
2
+ 5
19 , n
Ans.
Page : 5 of 15 TRIG. EQUATIONS
Self Practice Problems : 1.
Solve
cos2 ? (
2 + 1) cos -
1 2
= 0
2.
Solve
Ans.
Type - 3
4cos ? 3sec = tan
(1)
2n ? , n
3
or
2n ? , n
4
(2)
n + (? 1)n
where
=
sin?1
- 1- 8
17
,
n
or
n + (?1)n
where
=
sin?1
- 1+ 8
17
,
n
Trigonometric equations which can be solved by transforming a sum or difference of trigonometric ratios into their product.
Solved Example # 8 Solve cos3x + sin2x ? sin4x = 0
Solution.
cos3x + sin2x ? sin4x = 0 cos3x ? 2cos3x.sinx = 0
cos3x = 0
3x = (2n + 1) 2 , n
x = (2n + 1) 6 , n solution of given equation is
(2n + 1) 6 , n
or
cos3x + 2cos3x.sin(? x) = 0
cos3x (1 ? 2sinx) = 0
or
1 ? 2sinx = 0
1
or
sinx = 2
or
x = n + (?1)n 6 , n
n + (?1)n 6 , n
Ans.
Self Practice Problems :
1.
Solve sin7 = sin3 + sin
TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: 0 903 903 7779, 98930 58881 , BHOPAL
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2.
Solve 5sinx + 6sin2x +5sin3x + sin4x = 0
3.
Solve cos ? sin3 = cos2
Ans. (1) (2) (3)
n
3 ,n
or
n
2 ,n
or
2n 3 , n or
n 2 ? 12 , n
2 2n ? 3 , n
2n ? 2 , n
or
n + 4 , n
Type - 4 Trigonometric equations which can be solved by transforming a product of trigonometric ratios into their sum or difference.
Solved Example # 9
Solve sin5x.cos3x = sin6x.cos2x
Solution.
sin5x.cos3x = sin6x.cos2x
sin8x + sin2x = sin8x + sin4x
2sin2x.cos2x ? sin2x = 0
sin2x = 0
or
2cos2x ? 1 = 0
2x = n, n or
1 cos2x = 2
x = n , n or 2
2x = 2n ? 3 , n
x
=
n
?
6,
n
Solution of given equation is
Type - 5
n , n 2
or
n ?
6
, n
2sin5x.cos3x = 2sin6x.cos2x sin4x ? sin2x = 0 sin2x (2cos2x ? 1) = 0
Ans.
Trigonometric Equations of the form a sinx + b cosx = c, where a, b, c R, can be solved by dividing both sides of the equation by a2 + b2 .
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