OP Malhotra Class 11 Maths Solutions Chapter 6 Trigonometric Equations Ex 6

Access to comprehensive S Chand Class 11 Maths Solutions Chapter 6 Trigonometric Equations Ex 6 encourages independent learning.

S Chand Class 11 ICSE Maths Solutions Chapter 6 Trigonometric Equations Ex 6

Question 1.
Solve the following equations for 0 ≤ x ≤ 2π.
(i) 2 sin x – 1 = 0
(ii) sin x cos x = 0
(iii) tan θ + \(\sqrt{3}\) = 0
(iv) sin θ cos θ = \(\frac { 1 }{ 2 }\)
(v) 2 sin² θ = 3 cos θ
(vi) 2 + 7 tan² θ = 3.25 sec² θ
Solution:
(i) Given 2 sin x – 1 = θ ⇒ sin x = \(\frac { 1 }{ 2 }\) = sin \(\frac { π }{ 6 }\)

(ii) sin x cos x = 0 ; 0 ≤ x ≤ 2π
⇒ \(\frac { 1 }{ 2 }\) sin 2x = 0
⇒ sin 2x = 0
⇒ 2x = nπ
⇒ x = \(\frac { nπ }{ 2 }\); where n ∈ I
∴ x = 0, \(\frac { π }{ 2 }\), π, \(\frac { 3π }{ 2 }\), 2π

(iii) Given tan θ + \(\sqrt{3}\) = 0 ⇒ tan θ = – \(\sqrt{3}\) ; 0 ≤ θ ≤ 2π ⇒ tan θ = – tan \(\frac { π }{ 3 }\) = tan(-\(\frac { π }{ 3 }\))

(iv) sin θ cos θ = \(\frac { 1 }{ 2 }\) ⇒ 2 sin θ cos θ = 1 ⇒ sin 2θ = 1 = sin \(\frac { π }{ 2 }\)

Question 2.
cos θ + sin θ – sin 2θ = \(\frac { 1 }{ 2 }\), 0 < θ < \(\frac { π }{ 2 }\)
Solution:
Given sin θ + cos θ – sin 2θ = \(\frac { 1 }{ 2 }\), 0 < θ < \(\frac { π }{ 2 }\) ⇒ 2 (sin θ + cos θ) = 1 + sin 2θ
On squaring both sides ; we have
4 (sin θ + cos θ)² = (1 + 2 sin 2θ)²
⇒ 4 [sin² θ + cos² θ + sin 2θ] = [1 + 4 sin² 2θ + 4 sin 2θ]
⇒ 4 [1 + sin 2θ] = 1 + 4 sin² 2θ + 4 sin 2θ ⇒ 4 sin² 2θ = 3

Question 3.
sin 5θ = cos 2θ, 0° < θ < 180°.
Solution:

Question 4.
cot²θ – (1 + \(\sqrt{3}\))cot θ + \(\sqrt{3}\) = 0, 0 < θ < \(\frac { π }{ 2 }\)
Solution:
Given cot²θ – (1 + \(\sqrt{3}\))cot θ + \(\sqrt{3}\) = 0, 0 < θ < \(\frac { π }{ 2 }\)
⇒ cot²θ – cot θ – \(\sqrt{3}\) cot θ + \(\sqrt{3}\) = 0
⇒ cot θ (cot θ – 1) – \(\sqrt{3}\) (cot θ – 1) = 0
⇒ (cot θ – 1) (cot θ – \(\sqrt{3}\)) = 0
⇒ cot θ – 1 = 0 or cot θ – \(\sqrt{3}\) = 0
⇒ tan θ = 1 or tan θ = \(\frac{1}{\sqrt{3}}\)
⇒ θ = \(\frac { π }{ 4 }\) or θ = \(\frac { π }{ 6 }\)
Since 0 < θ < \(\frac { π }{ 2 }\)
∴ θ = \(\frac { π }{ 4 }\), \(\frac { π }{ 6 }\)

Question 5.
sin x + cos (x + 30°) = 0, 0° < x < 360°.
Solution:

Question 6.
cos 6θ + cos 4θ + cos 2θ + 1 = 0, 0° < θ < 180°
Solution:

Question 7.
sin 7θ + sin 4θ + sin θ = 0, 0 < θ < \(\frac { π }{ 2 }\)
Solution:

Solve, giving the general value.

Question 8.
2 cos² θ – 5 cos θ + 2 = 0
Solution:
2 cos² θ – 5 cos θ + 2 = 0
⇒ cos θ = \(\frac{5 \pm \sqrt{25-16}}{4}=\frac{5 \pm 3}{4}=2, \frac{1}{2}\)
since |cos θ | ≤ 1 ∴ cos θ = \(\frac { 1 }{ 2 }\) = cos \(\frac { π }{ 3 }\)
⇒ θ = 2nπ ± \(\frac { π }{ 3 }\); where n ∈ I
which gives the required general soln.

Question 9.
2 sin² x + \(\sqrt{3}\)cos x + 1 = 0
Solution:

Question 10.
2 + \(\sqrt{3}\)sec x – 4cos x = 2\(\sqrt{3}\)
Solution:

Question 11.
tan² θ – (1 + \(\sqrt{3}\))tan θ + \(\sqrt{3}\) = 0
Solution:
Given tan² θ – (1 + \(\sqrt{3}\))tan θ + \(\sqrt{3}\) = 0
⇒ tan θ (tan θ – 1) – \(\sqrt{3}\) (tan θ – 1) = 0
⇒ (tan θ – 1) (tan θ – \(\sqrt{3}\)) = 0
⇒ tan θ = 1 or tan θ = \(\sqrt{3}\)
∴ tan θ = \(\frac { π }{ 4 }\) or tan θ = tan\(\frac { π }{ 3 }\)
⇒ θ = nπ + \(\frac { π }{ 4 }\) or θ = mπ + \(\frac { π }{ 3 }\)
∴ θ = nπ + \(\frac { π }{ 4 }\) or mπ + \(\frac { π }{ 3 }\) where m, n ∈ I

Question 12.
tan θ + 4 cot 2θ + 1 = 0
Solution:
Given tan θ + 4 cot 2θ + 1 = 0
⇒ tan θ + \(\frac{4\left(1-\tan ^2 \theta\right)}{2 \tan \theta}\) + 1 = 0
⇒ 2 tan² θ + 4 – 4 tan² θ + 2 tan θ = 0
⇒ 2 tan² θ – 2 tan θ – 4 = 0
⇒ tan² θ – tan θ – 2 = 0
⇒ tan θ = \(\frac{-(-1) \pm \sqrt{1+8}}{2}=\frac{1 \pm 3}{2}\) = 2, – 1
Case-I. When tan θ = 2 ⇒ tan θ = tan α ⇒ θ = nπ + α, where α = tan^-1 2
Case-II. When tan θ = – 1 = – tan \(\frac { π }{ 4 }\) = tan(-\(\frac { π }{ 4 }\)) ⇒ θ = mπ – \(\frac { π }{ 4 }\) ; m ∈ I
Thus, θ = mπ – \(\frac { π }{ 4 }\), nπ + α ; where m, n ∈ I

Question 13.
tan θ + tan² θ + \(\sqrt{3}\) tan θ tan 2θ = \(\sqrt{3}\)
Solution:

Question 14.
cot θ + tan θ = 2 cosec θ
Solution:
Given cot θ + tan θ = 2 cosec θ … (1)
⇒ \(\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}=\frac{2}{\sin \theta}\)
⇒ \(\frac{\cos ^2 \theta+\sin ^2 \theta}{\cos \theta \sin \theta}=\frac{2}{\sin \theta}\)
⇒ \(\frac{2}{\sin 2 \theta}=\frac{2}{\sin \theta}\)
⇒ sin 2θ = sin θ
⇒ sin θ (2 cos θ – 1) = 0
But sin θ ≠ 0 otherwise eqn. (1) is not possible.
∴ 2 cos θ – 1 = 0⇒ cos θ = \(\frac { 1 }{ 2 }\) = cos \(\frac { π }{ 3 }\) ⇒ θ = 2nπ ± \(\frac { π }{ 3 }\), where n ∈ I

Question 15.
2 cos θ + cos 3θ = 0
Solution:

Question 16.
2 sin 2x – sin x = 0
Solution:

Question 17.
tan 2x + 2 tan x = 0
Solution:

Question 18.
sin 7θ + sin 4θ + sin θ = 0
Solution:

Question 19.
cos θ + cos 2θ + cos 3θ = 0
Solution:

Question 20.
sin θ + cos θ = \(\sqrt{2}\)
Solution:

Question 21.
sin θ + \(\sqrt{3}\) cos θ = \(\sqrt{2}\)
Solution:
Given sin θ + \(\sqrt{3}\) cos θ = \(\sqrt{2}\)
⇒ \(\frac { 1 }{ 2 }\) sin θ + \(\frac{\sqrt{3}}{2}\) cos θ = \(\frac{1}{\sqrt{2}}\)
⇒ sin θ cos \(\frac { π }{ 3 }\) + cos θ sin \(\frac { π }{ 3 }\) =sin \(\frac { π }{ 4 }\)
⇒ sin (θ + \(\frac { π }{ 3 }\)) = sin \(\frac { π }{ 4 }\)
⇒ θ + \(\frac { π }{ 3 }\) = nπ + (-1)ⁿ \(\frac { π }{ 4 }\)
⇒ θ = nπ – \(\frac { π }{ 3 }\) + (-1)ⁿ \(\frac { π }{ 4 }\), where n ∈ I

Question 22.
\(\sqrt{2}\) sec θ + tan θ = 1
Solution:

Question 23.
3 – 2 cos θ – 4 sin θ – cos 2θ + sin 2θ = 0
Solution:
Given 3 – 2 cos θ – 4 sin θ – cos 2θ + sin 2θ = 0
⇒ 3 – 2 cos θ – 4 sin θ – (1 – 2 sin² θ) + 2 sin θ cos θ = 0
⇒ 2 – 2 cos θ – 4 sin θ + 2 sin² θ + 2 sin θ cos θ = 0
⇒ 1 – cos θ – 2 sin θ + sin² θ + sin θ cos θ = 0
⇒ 1 – cos θ – sin θ – sin θ + sin² θ + sin θ cos θ = 0
⇒ (1 – sin θ) – cos θ(1 – sin θ) – sin θ(1 – sin θ) = 0
⇒ (1 – sin θ) (1 – cos θ – sin θ) = 0
∴ 1 – sin θ = 0 or 1 – cos θ – sin θ = 0
Case-I:
1 – sin θ = 0 ⇒ sin θ = 1 = sin \(\frac { π }{ 2 }\) ⇒ θ = nπ + (-1)ⁿ\(\frac { π }{ 2 }\) ; n ∈ I

Case-II:
1 – cos θ – sin θ = 0
⇒ cos θ + sin θ = 1
⇒ \(\frac{1}{\sqrt{2}}\) cos θ + \(\frac{1}{\sqrt{2}}\) sin θ = \(\frac{1}{\sqrt{2}}\)
⇒ cos θ cos \(\frac{\pi}{4}+\sin \theta \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}\)
⇒ cos \(\left(\theta-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}=\cos \frac{\pi}{4}\)
⇒ θ – \(\frac { π }{ 4 }\) = 2nπ ± \(\frac { π }{ 4 }\)
⇒ θ = 2nπ ± \(\frac { π }{ 4 }\) + \(\frac { π }{ 4 }\)
⇒ θ = 2nπ, 2nπ ± \(\frac { π }{ 2 }\)
Hence general soln. is given by θ = nπ + (-1)ⁿ, 2nπ, 2rm + \(\frac { π }{ 2 }\), n ∈ I

Question 24.
If the equation a cos 2θ + b sin 2θ = c has θ₁, θ₂ as its roots, prove that
(i) tan θ₁ + tan θ₂ = \(\frac{2 b}{c+a}\)
(i) tan θ₁.tan θ₂ = \(\frac{c-a}{c+a}\)
Solution:
(i) Given
a cos θ + b sin 2θ = c … (1)
⇒ \(\frac{a\left(1-\tan ^2 \theta\right)}{1+\tan ^2 \theta}+b\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\) = c
⇒ a (1 – tan² θ) + 2b tan θ = c (1 + tan² 0)
⇒ (a + c) tan² θ – 2b tan θ + c – a = 0 … (2)
since it is given that θ₁, θ₂ are the roots of eqn. (1)
∴ tan θ₁ and tan θ₂ are the roots of eqn. (2)
∴ tan θ₁ + tan θ₂ = sum of roots = – \(\left(\frac{-2 b}{a+c}\right)=\frac{2 b}{a+c}\)

(ii) product of roots = tan θ₁ tan θ₂ = \(\frac{c-a}{c+a}\)

Question 25.
If α, ß are two different values of θ lying between 0 and 2π which satisfy the equation 6 cos θ + 8 sin θ = 9, find the value of sin (α + ß).
Solution:

Question 26.
Find all the values of θ satisfying the equation cos 2θ – cos 8θ + cos 6θ = 1, such that 0 ≤ θ ≤ n.
Solution:

Question 27.
sec θ – cosec θ = \(\frac { 4 }{ 3 }\)
Solution:

Question 28.
Find the smallest positive number p for which the equation cos (p sin x) = sin (p cos x) has a solution when x ∈ [0, 2π].
Solution:
Given cos (p sin x) = sin (p cos x) x ∈ [0, π]

From case-I and case-II, we conclude that the smallest positive value of p for which given eqn. is \(\frac{\pi}{2 \sqrt{2}}\).