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 Page 1


11. Trigonometric Equations
Exercise 11.1
1 A. Question
Find the general solutions of the following equations :
i. 
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
we have,
We know that sin 30° = sin p/6 = 0.5
? 
? it matches with the form sin x = sin y
Hence,
 ,where n ? Z
1 A. Question
Find the general solutions of the following equations :
i. 
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
we have,
We know that sin 30° = sin p/6 = 0.5
? 
? it matches with the form sin x = sin y
Page 2


11. Trigonometric Equations
Exercise 11.1
1 A. Question
Find the general solutions of the following equations :
i. 
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
we have,
We know that sin 30° = sin p/6 = 0.5
? 
? it matches with the form sin x = sin y
Hence,
 ,where n ? Z
1 A. Question
Find the general solutions of the following equations :
i. 
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
we have,
We know that sin 30° = sin p/6 = 0.5
? 
? it matches with the form sin x = sin y
Hence,
 ,where n ? Z
1 B. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that, cos 150°  = 
? 
If cos x = cos y then x = 2n p ± y, where n ? Z.
For above equation y = 5p / 6
? x = 2np ± 5p / 6 ,where n ? Z
Thus, x gives the required general solution for the given trigonometric equation.
1 B. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that, cos 150°  = 
Page 3


11. Trigonometric Equations
Exercise 11.1
1 A. Question
Find the general solutions of the following equations :
i. 
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
we have,
We know that sin 30° = sin p/6 = 0.5
? 
? it matches with the form sin x = sin y
Hence,
 ,where n ? Z
1 A. Question
Find the general solutions of the following equations :
i. 
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
we have,
We know that sin 30° = sin p/6 = 0.5
? 
? it matches with the form sin x = sin y
Hence,
 ,where n ? Z
1 B. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that, cos 150°  = 
? 
If cos x = cos y then x = 2n p ± y, where n ? Z.
For above equation y = 5p / 6
? x = 2np ± 5p / 6 ,where n ? Z
Thus, x gives the required general solution for the given trigonometric equation.
1 B. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that, cos 150°  = 
? 
If cos x = cos y then x = 2n p ± y, where n ? Z.
For above equation y = 5p / 6
? x = 2np ± 5p / 6 ,where n ? Z
Thus, x gives the required general solution for the given trigonometric equation.
1 C. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that sin x, and cosec x have negative values in the 3
rd
 and 4
th
 quadrant.
While giving a solution, we always try to take the least value of y
The fourth quadrant will give the least magnitude of y as we are taking an angle in a clockwise sense (i.e.,
negative angle)
-v2 = -cosec (p/4) = cosec (-p/4) { ? sin -? = -sin ? }
? 
? 
If sin x = sin y ,then x = np + (– 1)
n
y , where n ? Z.
For above equation y = 
? x = np + (-1)
n
,where n ? Z
Or x = np + (-1)
n+1
 ,where n ? Z
Thus, x gives the required general solution for given trigonometric equation.
1 C. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
Page 4


11. Trigonometric Equations
Exercise 11.1
1 A. Question
Find the general solutions of the following equations :
i. 
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
we have,
We know that sin 30° = sin p/6 = 0.5
? 
? it matches with the form sin x = sin y
Hence,
 ,where n ? Z
1 A. Question
Find the general solutions of the following equations :
i. 
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
we have,
We know that sin 30° = sin p/6 = 0.5
? 
? it matches with the form sin x = sin y
Hence,
 ,where n ? Z
1 B. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that, cos 150°  = 
? 
If cos x = cos y then x = 2n p ± y, where n ? Z.
For above equation y = 5p / 6
? x = 2np ± 5p / 6 ,where n ? Z
Thus, x gives the required general solution for the given trigonometric equation.
1 B. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that, cos 150°  = 
? 
If cos x = cos y then x = 2n p ± y, where n ? Z.
For above equation y = 5p / 6
? x = 2np ± 5p / 6 ,where n ? Z
Thus, x gives the required general solution for the given trigonometric equation.
1 C. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that sin x, and cosec x have negative values in the 3
rd
 and 4
th
 quadrant.
While giving a solution, we always try to take the least value of y
The fourth quadrant will give the least magnitude of y as we are taking an angle in a clockwise sense (i.e.,
negative angle)
-v2 = -cosec (p/4) = cosec (-p/4) { ? sin -? = -sin ? }
? 
? 
If sin x = sin y ,then x = np + (– 1)
n
y , where n ? Z.
For above equation y = 
? x = np + (-1)
n
,where n ? Z
Or x = np + (-1)
n+1
 ,where n ? Z
Thus, x gives the required general solution for given trigonometric equation.
1 C. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that sin x, and cosec x have negative values in the 3
rd
 and 4
th
 quadrant.
While giving a solution, we always try to take the least value of y
The fourth quadrant will give the least magnitude of y as we are taking an angle in a clockwise sense (i.e.,
negative angle)
-v2 = -cosec (p/4) = cosec (-p/4) { ? sin -? = -sin ? }
? 
? 
If sin x = sin y ,then x = np + (– 1)
n
y , where n ? Z.
For above equation y = 
? x = np + (-1)
n
,where n ? Z
Or x = np + (-1)
n+1
 ,where n ? Z
Thus, x gives the required general solution for given trigonometric equation.
1 D. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that sec x and cos x have positive values in the 1
st
 and 4
th
 quadrant.
While giving a solution, we always try to take the least value of y
both quadrants will give the least magnitude of y.
We can choose any one, in this solution we are assuming a positive value.
? 
Page 5


11. Trigonometric Equations
Exercise 11.1
1 A. Question
Find the general solutions of the following equations :
i. 
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
we have,
We know that sin 30° = sin p/6 = 0.5
? 
? it matches with the form sin x = sin y
Hence,
 ,where n ? Z
1 A. Question
Find the general solutions of the following equations :
i. 
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
we have,
We know that sin 30° = sin p/6 = 0.5
? 
? it matches with the form sin x = sin y
Hence,
 ,where n ? Z
1 B. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that, cos 150°  = 
? 
If cos x = cos y then x = 2n p ± y, where n ? Z.
For above equation y = 5p / 6
? x = 2np ± 5p / 6 ,where n ? Z
Thus, x gives the required general solution for the given trigonometric equation.
1 B. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that, cos 150°  = 
? 
If cos x = cos y then x = 2n p ± y, where n ? Z.
For above equation y = 5p / 6
? x = 2np ± 5p / 6 ,where n ? Z
Thus, x gives the required general solution for the given trigonometric equation.
1 C. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that sin x, and cosec x have negative values in the 3
rd
 and 4
th
 quadrant.
While giving a solution, we always try to take the least value of y
The fourth quadrant will give the least magnitude of y as we are taking an angle in a clockwise sense (i.e.,
negative angle)
-v2 = -cosec (p/4) = cosec (-p/4) { ? sin -? = -sin ? }
? 
? 
If sin x = sin y ,then x = np + (– 1)
n
y , where n ? Z.
For above equation y = 
? x = np + (-1)
n
,where n ? Z
Or x = np + (-1)
n+1
 ,where n ? Z
Thus, x gives the required general solution for given trigonometric equation.
1 C. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that sin x, and cosec x have negative values in the 3
rd
 and 4
th
 quadrant.
While giving a solution, we always try to take the least value of y
The fourth quadrant will give the least magnitude of y as we are taking an angle in a clockwise sense (i.e.,
negative angle)
-v2 = -cosec (p/4) = cosec (-p/4) { ? sin -? = -sin ? }
? 
? 
If sin x = sin y ,then x = np + (– 1)
n
y , where n ? Z.
For above equation y = 
? x = np + (-1)
n
,where n ? Z
Or x = np + (-1)
n+1
 ,where n ? Z
Thus, x gives the required general solution for given trigonometric equation.
1 D. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that sec x and cos x have positive values in the 1
st
 and 4
th
 quadrant.
While giving a solution, we always try to take the least value of y
both quadrants will give the least magnitude of y.
We can choose any one, in this solution we are assuming a positive value.
? 
If cos x = cos y then x = 2n p ± y, where n ? Z.
For above equation y = p / 4
? x = 2np ±  ,where n ? Z
Thus, x gives the required general solution for the given trigonometric equation.
1 D. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.
Given,
We know that sec x and cos x have positive values in the 1
st
 and 4
th
 quadrant.
While giving a solution, we always try to take the least value of y
both quadrants will give the least magnitude of y.
We can choose any one, in this solution we are assuming a positive value.
? 
If cos x = cos y then x = 2n p ± y, where n ? Z.
For above equation y = p / 4
? x = 2np ±  ,where n ? Z
Thus, x gives the required general solution for the given trigonometric equation.
1 E. Question
Find the general solutions of the following equations :
Answer
Ideas required to solve the problem:
The general solution of any trigonometric equation is given as –
• sin x = sin y, implies x = np + (– 1)
n
y, where n ? Z.
• cos x = cos y, implies x = 2n p ± y, where n ? Z.
• tan x = tan y, implies x = n p + y, where n ? Z.

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