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

Students can cross-reference their work with S Chand Class 11 Maths Solutions Chapter 6 Trigonometric Equations Chapter Test to ensure accuracy.

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

Question 1.
Find the general value of 6 which satisfies the equation sin² θ = \(\frac { 3 }{ 4 }\).
Solution:
Given \(\sin ^2 \theta=\frac{3}{4}=\left(\frac{\sqrt{3}}{2}\right)^2=\sin ^2 \frac{\pi}{3}\)
⇒ θ = nπ ± \(\frac { π }{ 3 }\) ; where n ∈ I [∵ sin² θ = sin² α ⇒ θ = nπ ± [ \(\frac { π }{ 3 }\); n ∈ I]

Question 2.
Find the general solution of sin 2θ = cos θ.
Solution:
Given sin 2θ = cos θ = sin (\(\frac { π }{ 2 }\) – θ)
⇒ 2θ = nπ + (-1)ⁿ(\(\frac { π }{ 2 }\) – θ)
⇒ θ = \(\frac { nπ }{ 2 }\)+(-1)ⁿ(\(\frac { π }{ 4 }\) – \(\frac { θ }{ 2 }\))
Aliter: sin 2θ = cos θ ⇒ 2 sin θ cos θ – cos θ = θ ⇒ cos θ (2 sin θ – 1) = 0
cos θ = 0 or 2 sin θ – 1 = 0
⇒ q = nπ + \(\frac { π }{ 2 }\) ⇒ sin θ = \(\frac { 1 }{ 2 }\) = sin \(\frac { π }{ 6 }\)
⇒ θ = nπ + (- 1)ⁿ \(\frac { π }{ 6 }\) ; n ∈ I
Hence θ = nπ + \(\frac { π }{ 2 }\), nπ + (-1)ⁿ\(\frac { π }{ 6 }\) ; where n ∈ I

Question 3.
Solve: sin 5x – sin 3x – sin x = 0, 0° < x < 360°.
Solution:
Given sin 5x – sin 3x – sin x = 0, 0° < x < 360°
⇒ \(2 \cos \left(\frac{5 x+3 x}{2}\right) \sin \left(\frac{5 x-3 x}{2}\right)-\sin x\)
⇒ 2 cos 4x sin x – sin x = 0
⇒ sin x(2cos 4x – 1) = 0
∴ sin x = 0 or 2 cos 4x – 1 = 0

Case-I: sin x = 0 ; 0 < x < 360° ⇒ x = 0°, π, 2π but 0 < x < 360° ∴ x = π

Case-II: When 2 cos 4x – 1 = 0 ⇒ cos 4x = \(\frac { 1 }{ 2 }\) = cos \(\frac { π }{ 3 }\)
⇒ 4x = 2nπ + \(\frac { π }{ 3 }\)
⇒ x = \(\frac { nπ }{ 2 }\) ± \(\frac { π }{ 12 }\) ; n ∈ I
∴ x = \(\frac{\pi}{12}, \frac{\pi}{2} \pm \frac{\pi}{12}, \pi \pm \frac{\pi}{12}, \frac{3 \pi}{12} \pm \frac{\pi}{12}, 2 \pi-\frac{\pi}{12}\)
i.e. x = \(\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{7 \pi}{12}, \frac{11 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}, \frac{19 \pi}{12}, \frac{23 \pi}{12}\)
Thus x = π, \(\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{7 \pi}{12}, \frac{11 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}, \frac{19 \pi}{12}, \frac{23 \pi}{12}\)

Question 4.
Find the general solution of tan² θ = \(\frac { 1 }{ 3 }\), and hence find those values of θ for which – π ≤ θ ≤ π.
Solution:

Question 5.
Solve : \(\tan \left(\frac{\pi}{4}+\theta\right)+\tan \left(\frac{\pi}{4}-\theta\right)\) = 4
Solution:

Question 6.
Solve the equation \(\sqrt{3}\) cos x + sin x = 1 for – 2π < x < 2π.
Solution: