Referring to the AP Inter 1st Year Maths Study Material Chapter Trigonometric Functions Exercise 3e Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Trigonometric Functions Exercise 3e
I. Find the principal and general solutions of the following equations.
Question 1.
tan x = √3
Solution:
We know that tan \(\frac{1}{2}\) = √3 and tan\(\left(\frac{4 \pi}{3}\right)\) = tan\(\left(\pi+\frac{\pi}{3}\right)\)
= tan\(\frac{\pi}{3}\) = √3
The principal solutions are x = \(\frac{\pi}{3}\) and \(\frac{4 \pi}{3}\)
tan x = tan\(\frac{\pi}{3}\) ⇒ x = nπ + \(\frac{\pi}{3}\) where n ∈ Z
The general solution is x = nπ + \(\frac{\pi}{3}\) where n ∈ Z
Question 2.
sec x = 2
Solution:
We know that sec\(\frac{\pi}{3}\) = 2 and sec \(\frac{5 \pi}{3}\)
= sec (2π – \(\frac{\pi}{3}\)) = sec \(\frac{\pi}{3}\) = 2
The principal solutions are x = \(\frac{\pi}{3}\) and \(\frac{5 \pi}{3}\)
sec x = sec \(\frac{\pi}{3}\) ⇒ cos x = cos \(\frac{\pi}{3}\) [∵ sec x = \(\frac{1}{\cos x}\)]
⇒ x – 2nπ ± \(\frac{\pi}{3}\) where n ∈ Z
∴ The general solution is x = 2nπ ± \(\frac{\pi}{3}\); where n ∈ Z
Question 3.
cot x = √3
Solution:
We know that, cot \(\frac{\pi}{3}\) = √3
cot (π – \(\frac{\pi}{6}\)) = -cot\(\frac{\pi}{6}\) = -√3 and cot(2π – \(\frac{\pi}{6}\))
= -cot\(\frac{\pi}{6}\) = -√3
⇒ cot \(\frac{5 \pi}{6}\) = -√3 and cot \(\frac{11 \pi}{6}\) = -√3
The principal solutions are x = \(\frac{5 \pi}{6}\) and \(\frac{11 \pi}{6}\)
cot x = cot \(\frac{5 \pi}{6}\) ⇒ tanx = tan\(\frac{5 \pi}{6}\)
⇒ x = nx + \(\frac{5 \pi}{6}\) , where n ∈ Z
∴ The general solution is x = nx + \(\frac{5 \pi}{6}\), where n ∈ Z
Question 4.
cosec x = -2
Solution:
We know that, cosec \(\frac{\pi}{6}\) = 2
cosec (π + \(\frac{\pi}{6}\)) = – cosec \(\frac{\pi}{6}\) = -2 and cosec (2π – \(\frac{\pi}{6}\)) = -cosec \(\frac{\pi}{6}\) = -2
⇒ cosec \(\frac{7 \pi}{6}\) = -2 and cosec \(\frac{11 \pi}{6}\) = -2
The principal solutions are x = \(\frac{7 \pi}{6}\) and \(\frac{11 \pi}{6}\)
cosec x = cosec \(\frac{7 \pi}{6}\) ⇒ sinx = sin\(\frac{7 \pi}{6}\)
x = nπ + (-1)n\(\frac{7 \pi}{6}\)
∴ The general solution is x = nπ + (-1)n\(\frac{7 \pi}{6}\) where n ∈ Z
II. Find the general solution for each of the following equations.
Question 1.
cos 4x = cos 2x
Solution:
Given that, cos 4x = cos 2x ⇒ cos 4x – cos 2x = 0
⇒ -2 sin\(\left(\frac{4 x+2 x}{2}\right)\)sin\(\left(\frac{4 x-2 x}{2}\right)\) = 0
⇒ sin 3x sin x = 0
⇒ sin 3x = 0 or sin x = 0
3x = nπ or x = nπ, where n ∈ Z ⇒ x = \(\frac{\mathrm{n} \pi}{3}\) or x = nx, where n ∈ Z
Question 2.
cos 3x + cos x – cos 2x = 0.
Solution:
Given that cos 3x + cos x – cos 2x = 0
⇒ 2cos 2x cos x -cos 2x = 0
⇒ cos 2x (2 cos x – 1) = 0
⇒ cos 2x = 0 or 2cos x – 1 = 0
⇒ cos 2x = 0 or cos x = \(\frac{1}{2}\)
⇒ cos\(\frac{\pi}{3}\) ⇒ x = (2n + 1)\(\frac{\pi}{4}\) or x = 2nx ± \(\frac{\pi}{3}\), where n ∈ Z
Question 3.
sin 2x + cos x = 0
Solution:
Given that sin 2x + cos x = 0
⇒ 2 sin x cos x + cos x = 0
⇒ cos x (2 sin x + 1) = 0
⇒ cos x = 0 or 2 sin x + 1 = 0
cos x = 0 ⇒ x = (2n + 1)\(\frac{\pi}{2}\), where n ∈ Z
2 sin x + 1 = 0
⇒ sin x = \(\frac{-1}{2}\) = -sin\(\frac{\pi}{2}\) = sin(π + \(\frac{\pi}{2}\)) = sin \(\frac{7 \pi}{2}\)
⇒ x = nπ + (-1)n\(\frac{7 \pi}{6}\), where n ∈ Z
∴ The general solution is (2n + 1)\(\frac{\pi}{2}\) or nπ + (-1)n\(\frac{7 \pi}{2}\), n ∈ Z
Question 4.
sec2 2x = 1 – tan 2x.
Solution:
Given that 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 -1
⇒ 2x = nπ, where n ∈ Z
⇒ x = \(\frac{\mathrm{n} \pi}{2}\), where n ∈ Z
tan 2x = -1 = -tan \(\frac{\pi}{4}\) = tan (π – \(\frac{\pi}{4}\)) = tan \(\frac{3 \pi}{2}\)
⇒ 2x = nπ + \(\frac{3 \pi}{4}\), where n ∈ Z
⇒ x = \(\frac{\mathrm{n} \pi}{2}+\frac{3 \pi}{8}\), where n ∈ Z
∴ The general solution is \(\frac{\mathrm{n} \pi}{2}\) or \(\frac{\mathrm{n} \pi}{2}+\frac{3 \pi}{8}\), n ∈ Z
Question 5.
sin x + sin 3x + sin 5x = 0
Solution:
We know that, sin A + sin B = 2sin\(\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\) cos \(\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\)
⇒ sin x + sin 5x = 2 sin \(\left(\frac{6 \mathrm{x}}{2}\right)\) cos \(\left(\frac{-4 x}{2}\right)\) = 2 sin 3x cos 2x
⇒ 2sin 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
⇒ 3x = nπ
⇒ x = \(\frac{\mathrm{n} \pi}{3}\)
2 cos 2x + 1 = 0
⇒ cos 2x = –\(\frac{1}{2}\)
⇒ 2x = 2nπ ± \(\frac{2 \pi}{3}\)
⇒ x = nπ ± \(\frac{\pi}{3}\)
∴ The general solution is \(\frac{\mathrm{n} \pi}{3}\) or nπ ± \(\frac{\pi}{3}\), where n ∈ Z .