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STUDY PACKAGE
Subject : Mathematics Topic: Trigonometric Equation & Properties & Solution of Triangle
Index
?
1. Theory
2. Short Revision
3. Exercise
4. Assertion & Reason
5. Que. from Compt. Exams
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MATHEMATICS
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Page : 2 of 29
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
=
,
4
3 ,
4
9 ,
4
11 , ...........
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 GROUP MATHS BY SUHAAG SIR PH: (0755)- 32 00 000, 98930 58881
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)
2
,n
= (2n + 1)
2
,
n
= 2n,
n
= (2n + 1), n
= n,
n
Solved Example # 1
Solve
sin =
3 .
2
Solution.
Page : 3 of 29
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 GROUP MATHS BY SUHAAG SIR PH: (0755)- 32 00 000, 98930 58881
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
Ans. (1)
= n ?
4
,
n
(2)
2n 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 29
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)
2
,
n
or
n ?
4
,
n
Types of Trigonometric Equations :
TEKO CLASSES GROUP MATHS BY SUHAAG SIR PH: (0755)- 32 00 000, 98930 58881
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 6 , n
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
cosx =
-2+ 5
19
cosx = cos,
where
cos =
-2+ 5
19
x = 2n ? where
= cos?1
-
2
+ 5
19 , n
Ans.
Page : 5 of 29
Self Practice Problems : 1.
Solve
cos2 ? (
2 + 1) cos -
1 2
= 0
2.
Solve 4cos ? 3sec = tan
Ans. (1)
2n ? , n 3
or
2n ? , n
4
(2)
n + (? 1)n
where
=
sin?1
-
1- 8
17
,
n
Type - 3
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.
TEKO CLASSES GROUP MATHS BY SUHAAG SIR PH: (0755)- 32 00 000, 98930 58881
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
Self Practice Problems :
1.
Solve sin7 = sin3 + sin
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.
2.
Solve 5sinx + 6sin2x +5sin3x + sin4x = 0
3.
Solve cos ? sin3 = cos2
n
Ans. (1)
3 ,n
or
n 2
?
12
,
n
(2)
n 2
,
n
or
2 2n ? 3 , n
2n
(3)
3 , n or
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
1
2x = n, n or
cos2x = 2
x
=
n 2
,
n
or
2x = 2n ? 3 , n
Type - 5
x = n ? 6 , n Solution of given equation is
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 .
Page : 6 of 29
TEKO CLASSES GROUP MATHS BY SUHAAG SIR PH: (0755)- 32 00 000, 98930 58881
Solved Example # 10
Solve sinx + cosx = 2
Solution.
sinx + cosx = 2
Here a = 1, b = 1.
..........(i)
divide both sides of equation (i) by 2 , we get
1
1
sinx .
+ cosx. = 1
2
2
sinx.sin + cosx.cos = 1
4
4
cos
x -
4
=
1
x ? 4 = 2n, n
x = 2n +
4
,
n
Solution of given equation is
2n +
4
,
n
Ans.
Note : Trigonometric equation of the form a sinx + b cosx = c can also be solved by changing sinx and cosx into their corresponding tangent of half the angle.
Solved Example # 11
Solve 3cosx + 4sinx = 5
Solution.
Let
3cosx + 4sinx = 5
.........(i)
1- tan2 x
cosx =
2 1+ tan2 x
&
2
equation (i) becomes
2 tan x
sinx
=
2 1+ tan2 x
2
3
1- tan2 1+ tan2
x
2
x 2
+
4
2 tan x
2
1+
tan2
x 2
=
5
x tan 2 = t equation (ii) becomes
........(ii)
3
1 - 1+
t2 t2
+ 4
2 t 1+ t2
= 5
4t2 ? 4t + 1 = 0
(2t ? 1)2 = 0
1 t =
2
x t = tan
2
x 1 tan 2 = 2
x
1
tan 2 = tan, where tan = 2
x 2 = n +
x = 2n + 2
where
=
tan?1
1 2
,
n
Ans.
Self Practice Problems :
1.
Solve 3 cosx + sinx = 2
Page : 7 of 29
x
2.
Solve sinx + tan 2 = 0
Ans. (1)
2n + 6 , n
(2) x = 2n, n
Type - 6
Trigonometric equations of the form P(sinx ? cosx, sinx cosx) = 0, where p(y, z) is a polynomial, can be solved by using the substitution sinx ? cosx = t.
Solved Example # 12
TEKO CLASSES GROUP MATHS BY SUHAAG SIR PH: (0755)- 32 00 000, 98930 58881
Solve sinx + cosx = 1 + sinx.cosx
Solution.
Let
sinx + cosx = 1 + sinx.cosx
sinx + cosx = t sin2x + cos2x + 2 sinx.cosx = t2
........(i)
sinx.cosx = t2 -1
2
Now put
sinx + cosx = t and sinx.cosx = t2 -1 in (i), we get 2
t = 1 + t2 -1
2
t2 ? 2t + 1 = 0
t = 1
sinx + cosx = 1
t = sinx + cosx
.........(ii)
divide both sides of equation (ii) by 2 , we get
1
1
1
sinx. + cosx. =
2
2
2
cos x - 4
=
cos
4
x ?
4
= 2n ?
4
(i)
if we take positive sign, we get
x = 2n + 2 , n Ans.
(ii) if we take negative sign, we get
x = 2n, n
Ans.
Self Practice Problems:
1.
Solve sin2x + 5sinx + 1 + 5cosx = 0
2.
Solve 3cosx + 3sinx + sin3x ? cos3x = 0
3.
Solve
Ans.
Type - 7
(1 ? sin2x) (cosx ? sinx) = 1 ? 2sin2x.
(1)
n ? 4 , n
(2)
n ? 4 , n
(3)
2n +
2
,
n
or
2n, n
or
n +
4
,
n
Trigonometric equations which can be solved by the use of boundness of the trigonometric ratios sinx and cosx. Solved Example # 13
Solve
sinx cos
x 4
- 2 sin x
+
1 +
sin
x 4
- 2cos x cos x
= 0
Solution.
sinx cos
x - 2 sin x
4
+
1+ sin x - 2cos x cos x
4
= 0
.......(i)
x
x
sinx.cos ? 2sin2x + cosx + sin .cosx ? 2cos2x = 0
4
4
sin x.cos
x 4
+ sin x .cos x
4
?
2
(sin2x
+
cos2x)
+
cosx
=
0
sin 5x + cosx = 2
4
........(ii)
Page : 8 of 29
TEKO CLASSES GROUP MATHS BY SUHAAG SIR PH: (0755)- 32 00 000, 98930 58881
Now equation (ii) will be true if
5 x sin = 1
4
and cosx = 1
5 x 4
= 2n +
2
,
n
and
x = 2m, m
x
=
(8n
+ 5
2)
,
n
........(iii)
and
x = 2m, m
Now to find general solution of equation (i)
(8n + 2) 5
= 2m
8n + 2 = 10m
5m - 1
n= 4
if
m = 1
if
m = 5
.........
.........
if
m = 4p ? 3, p
then then ......... ......... then
n = 1 n = 6 .........
......... n = 5p ? 4, p
........(iv)
general solution of given equation can be obtained by substituting either m = 4p ? 3 in
equation (iv) or n = 5p ? 4 in equation (iii)
general solution of equation (i) is
(8p ? 6), p
Ans.
Self Practice Problems :
1.
Solve sin3x + cos2x = ? 2
2.
Solve 3 sin 5x - cos2 x - 3 = 1 ? sinx
Ans. (1)
(4p ? 3)
2
,
p
(2)
2m +
2
,
m
Exercise -1
Part : (A) Only one correct option
1.
The solution set of the equation 4sin.cos ? 2cos ? 2 3 sin + 3 = 0 in the interval (0, 2) is
(A)
3 4
,
7 4
(B)
3
,
5 3
(C)
3 4
,
,
, 3
5 3
(D)
6
,
5 , 6
11 6
2.
All solutions of the equation, 2 sin + tan = 0 are obtained by taking all integral values of m and n in:
2 (A) 2n + 3 , n
2 (B) n or 2m ? 3 where n, m
(C) n or m ? 3 where n, m
(D) n or 2m ? 3 where n, m
7
3.
If 20 sin2 + 21 cos - 24 = 0 & 4 < < 2 then the values of cot 2 is:
(A) 3
15 (B) 3
15 (C) - 3
4.
The general solution of sinx + sin5x = sin2x + sin4x is:
(A) 2 n ; n
(B) n ; n
(C) n/3 ; n
(D) - 3 (D) 2 n/3 ; n
1
5.
A triangle ABC is such that sin(2A + B) = . If A, B, C are in A.P. then the angle A, B, C are
2
respectively.
(A)
5 12
,
4
,
3
(B)
4
,
3,
5 12
(C)
3,
4
,
5 12
(D)
3,
5 12
,
4
................
................
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