Class 11 - Mathematics
Trigonometric Function - Exercise 3.4
Top Block 1
Question 1:
Find the principal and general solutions of the equation tan x = √3
Answer:
Given, tan x = √3
We know that tan π/3 = √3
and tan 4π/3 = tan (π + π/3) = tan π/3 = √3
Therefore, the principal solutions are: x = π/3 and x = 4π/3
Therefore, the general solution is
Now, tan x = tan π/3
⇒ x = nπ + π/3, where n є Z
Hence, the general solution is x = nπ + π/3, where n є Z
Question 2:
Find the principal and general solutions of the equation sec x = 2
Answer:
Given, sec x = 2
We know that sec π/3 = 2
and sec 5π/3 = tan (2π – π/3) = sec π/3 = 2
Therefore, the principal solutions are: x = π/3 and x = 5π/3
Now, sec x = sec π/3
⇒ cos x = cos π/3
⇒ x = 2nπ ± π/3, where n є Z
Hence, the general solution is x = 2nπ ± π/3, where n є Z
Question 3:
Find the principal and general solutions of the equation cot x = -√3
Answer:
Given, cot x = -√3
We know that cot π/6 = √3
Now, cot (π – π/6) = -cot π/6 = -√3
and cot (2π – π/6) = -cot π/6 = -√3
So, cot 5π/6 = -√3 and cot 11π/6 = -√3
Therefore, the principal solutions are: x = 5π/3 and x = 11π/3
Now, cot x = cot 5π/6
⇒ tan x = tan 5π/6
⇒ x = nπ + 5π/6, n є Z
Therefore, the general solution is x = nπ + 5π/6, n є Z
Question 4:
Find the general solution of cosec x = –2
Answer:
Given, cosec x = –2
We know that cosec π/6 = 2
Now, cosec (π + π/6) = – cosec π/6 = -2
and cosec (2π – π/6) = – cosec π/6 = -2
So, cosec 7π/6 = -2 and cosec 11π/6 = -2
Therefore, the principal solutions are: x = 7π/6 and x = 11π/6
Now, cosec x = cosec 7π/6
⇒ sin x = sin 7π/6
⇒ x = nπ + (-1)n 7π/6, n є Z
Therefore, the general solution is x = nπ + (-1)n 7π/6, n є Z
Question 5:
Find the general solution of the equation cos 4x = cos 2x
Answer:
Given, cos 4x = cos 2x
⇒ cos 4x – cos 2x = 0
⇒ -2 * sin {(4x + 2x)/2} * sin {(4x – 2x)/2} = 0 [Apply cos A – cos B formula]
⇒ sin 3x * sin x = 0
⇒ sin 3x = 0 or sin x = 0
⇒ 3x = nπ or x = nπ, where n є Z
⇒ x = nπ/3 or x = nπ, where n є Z
Question 6:
Find the general solution of the equation cos 3x + cos x – cos 2x = 0
Answer:
Given, cos 3x + cos x – cos 2x = 0
⇒ 2 * cos {(3x + x)/2} * cos {(3x – x)/2} – cos 2x = 0 [Apply cos A + cos B formula]
⇒ 2 * cos 2x * cos x – cos 2x = 0
⇒ cos 2x(2cos x – 1) = 0 ⇒ cos 2x = 0 or 2cos x – 1 = 0
⇒ cos 2x = 0 or cos x = 1/2
⇒ 2x = (2n + 1)π/2 or x = 2nπ ± π/3, where n є Z
⇒ x = (2n + 1)π/4 or x = 2nπ ± π/3, where n є Z
Mddle block 1
Question 7:
Find the general solution of the equation sin 2x + cos x = 0
Answer:
Given, sin 2x + cos x = 0
⇒ 2 * sin x * cos x + cos x = 0
⇒ cos x(2sin x + 1) = 0
⇒ cos x = 0 or 2sin x + 1 = 0
Now, cos x = 0 = cos π/2
⇒ x = (2n + 1)π/2, where n є Z
Again 2sin x + 1 = 0
⇒ sin x = -1/2 = -sin π/26
⇒ sin x = sin (π + π/6)
⇒ sin x = sin 7π/6
⇒ x = nπ + (-1)n 7π/6, where n є Z
Thus, the general solution is: = (2n + 1)π/2 or nπ + (-1)n 7π/6, where n є Z
Question 8:
Find the general solution of the equation sec2 2x = 1 – tan 2x
Answer:
Given, sec2 2x = 1 – tan 2x
⇒ 1 + tan2 2x = 1 – tan 2x
⇒ tan2 2x + tan 2x = 0
⇒ tan 2x(tan 2x + 1) = 0
⇒ tan 2x = 0 or tan 2x + 1 = 0
⇒ tan 2x = 0 or tan 2x = -1
⇒ tan 2x = tan 0 or tan 2x = -tan π/4
⇒ tan 2x = tan 0 or tan 2x = tan (π – π/4)
⇒ tan 2x = tan 0 or tan 2x = tan 3π/4
⇒ 2x = nπ + 0 or 2x = nπ + 3π/4, n є Z
⇒ x = nπ/2 or x = nπ/2 + 3π/8, n є Z
Thus, the general solution is nπ/2 or nπ/2 + 3π/8, n є Z
Question 9:
Find the general solution of the equation sin x + sin 3x + sin 5x = 0
Answer:
Given, sin x + sin 3x + sin 5x = 0 ⇒ sin x + sin 5x + sin 3x = 0
⇒ 2*sin(x + 5x)/2 * cos(x – 5x)/2 + sin 3x = 0
⇒ 2*sin 3x * cos(-2x) + sin 3x = 0
⇒ 2*sin 3x * cos 2x + sin 3x = 0
⇒ sin 3x(2 cos 2x + 1) = 0
⇒ sin 3x = 0 or 2 cos 2x + 1 = 0
⇒ sin 3x = 0 or cos 2x = -1/2
⇒ sin 3x = 0 or cos 2x = -1/2
⇒ sin 3x = 0 or cos 2x = -cos π/3
⇒ sin 3x = 0 or cos 2x = cos (π – π/3)
⇒ sin 3x = 0 or cos 2x = cos 2π/3
⇒ 3x = nπ or 2x = 2nπ ± 2π/3, n є Z ⇒ x = nπ/3 or x = nπ ± π/3
Therefore, the general solution is: nπ/3 or nπ ± π/3, n є Z
Bottom Block 3
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