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RD Sharma Solutions Commerce Maths Class 11 Class 11 Solutions Chapter - Trigonometric
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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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Document Description: RD Sharma Solutions Commerce Maths Class 11 Class 11 Solutions Chapter - Trigonometric for Commerce 2026 is part of Mathematics (Maths) Class 11 preparation. The notes and questions for RD Sharma Solutions Commerce Maths Class 11 Class 11 Solutions Chapter - Trigonometric have been prepared according to the Commerce exam syllabus. Information about RD Sharma Solutions Commerce Maths Class 11 Class 11 Solutions Chapter - Trigonometric covers topics like and RD Sharma Solutions Commerce Maths Class 11 Class 11 Solutions Chapter - Trigonometric Example, for Commerce 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions Commerce Maths Class 11 Class 11 Solutions Chapter - Trigonometric.
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