Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.8 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.8

Question 1.

Find the principal solution and general solutions of the following

(i) sin θ = – \(\frac{1}{\sqrt{2}}\)

(ii) cot θ = √3

(iii) tan θ = –\(\frac{1}{\sqrt{3}}\)

Answer:

(i) sin θ = – \(\frac{1}{\sqrt{2}}\)

We know that principal of sin θ lies in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)

sin θ = – \(\frac{1}{\sqrt{2}}\) < 0

∴ The principal value of sin θ lies in the IV quadrant.

sin θ = – \(\frac{1}{\sqrt{2}}\)

= – sin \(\left(\frac{\pi}{4}\right)\)

sin o = sin \(\left(-\frac{\pi}{4}\right)\)

Hence θ = \(-\frac{\pi}{4}\) is the principal solution.

The general solution is

θ = nπ + (- 1)^{n} . \(\left(-\frac{\pi}{4}\right)\) , n ∈ Z

θ = nπ + (- 1)^{n + 1} . \(\frac{\pi}{4}\) , n ∈ Z

(ii) cot θ = √3

The principal value of tan θ lies in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

Since tan θ = \(\frac{1}{\sqrt{3}}\) > 0

The principal value of tan θ lies in the I quadrant.

The general solution of tan θ is

θ = nπ + \(\frac{\pi}{6}\) , n ∈ Z

(iii) tan θ = –\(\frac{1}{\sqrt{3}}\)

The principal value of tan θ lies in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

Since tan θ = – \(\frac{1}{\sqrt{3}}\) > 0

The principal value of tan θ lies in the IV quadrant.

The general solution of tan θ is

θ = nπ – \(\frac{\pi}{6}\) , n ∈ Z

Question 2.

Solve the following equations for which solutions lies in the interval 0° ≤ 9 < 360°

(i) sin^{4}x = sin^{2}x

Answer:

sin^{4}x – sin^{2}x = 0

sin^{2} x (sin^{2} x – 1) = 0

sin^{2} x [ – (1 – sin^{2} x)] = 0

sin^{2}x × – cos^{2}x = 0

– sin^{2}x cos^{2}x = 0

(sin x cos x)^{2} = 0

(\(\frac { 1 }{ 2 }\) × 2 sin cos x)^{2} = 0

\(\frac { 1 }{ 4 }\) sin 2x = 0

sin 2x = 0

The general solution is

2x = nπ, n ∈ Z

x = \(\frac{\mathrm{n} \pi}{2}\), n ∈ Z

(ii) 2 cos^{2}x + 1 = – 3 cos x

Answer:

2 cos^{2}x + 1 = – 3 cos x

2 cos^{2}x + 3 cos x + 1 = 0

2 cos^{2}x + 2 cos x + cos x + 1 = 0

2 cos x (cos x + 1) + 1 (cos x + 1) = 0

(2 cos x + 1) (cos x + 1) = 0

2 cos x + 1 = 0 or cos x + 1 = 0

cos x = \(-\frac{1}{2}\) or cos x = – 1

To find the solution of cos x = \(-\frac{1}{2}\)

cos x = \(-\frac{1}{2}\)

To find the solution of cos x = – 1

cos x = – 1

cos x = cos π

The general solution is

x = 2nπ ± π, n ∈ Z

x = 2nπ + π or x = 2nπ – π, n ∈ Z

Consider x = 2nπ + π

when n = 0 , x = 0 + π = π ∈ (0°, 360°)

when n = 1 , x = 2π + π = 3π ∉ (0°, 360°)

Consider x = 2nπ – π

when n = 0, x = 0 – π ∉ (0°, 360°)

when n = 1, x = 2π – π = π ∈ (0°, 360°)

when n = 2, x = 4π – π = 3π ∉ (0°, 360°)

∴ The required solution are x = \(\frac{2 \pi}{3}\), \(\frac{4 \pi}{3}\), π

(iii) 2 sin^{2}x + 1 = 3 sin x

Answer:

2 sin^{2}x – 3 sin x + 1 = 0

2 sin^{2}x – 2 sin x – sin x + 1 = 0

2 sin x (sin x – 1) – 1 (sin x – 1) = 0

(2 sin x – 1)(sin x – 1) = 0

2 sin x – 1 = 0 or sin x – 1 = 0

sin x = \(\frac { 1 }{ 2 }\) or sin x = 1

To find the solution of sin x = \(\frac { 1 }{ 2 }\)

sin x = \(\frac { 1 }{ 2 }\)

sin x = sin \(\left(\frac{\pi}{6}\right)\)

The general solution is x = nπ + (-1)^{n}\(\frac{\pi}{6}\), n ∈ z

(iv) cos^{2}x = 1 – 3 sin x

Answer:

1 – 2 sin^{2}x = 1 – 3 sinx

2 sin^{2} x – 3 sin x = 0

sin x(2 sin x – 3) = 0smx =

sin x = 0 or 2 sin x – 3 = 0

sin x = 0 or sin x = \(\frac{3}{2}\)

sin x = \(\frac{3}{2}\) is not possible since sin x ≤ 1

∴ sin x = 0 = sin 0

The general solution is x = nit ,

When n = 0, x = 0 ∉ (0°, 360°)

When n = 1, x = π ∈ (0°, 360°)

When n = 2, x = 2π ∉ (0°, 360°)

∴ The required solutions is x = π

Question 3.

Solve the following equations:

(i) sin 5x – sin x = cos 3x

Answer:

2 cos 3 x . sin 2x = cos 3 x

2 cos 3x . sin 2x – cos3x = 0

cos 3x (2 sin 2x – 1) = 0

cos 3x = 0 or 2 sin 2x – 1 = 0

cos 3x = 0 or sin 2x = \(\frac { 1 }{ 2 }\)

To find the general solution of cos 3x = 0

The general solution of cos 3x = 0 is

3x = (2n + 1)\(\frac{\pi}{2}\), n ∈ Z

x = (2n + 1)\(\frac{\pi}{6}\), n ∈ Z

To find the general solution of sin 2x = \(\frac{1}{2}\)

sin 2x = \(\frac{1}{2}\)

sin 2x = sin \(\left(\frac{\pi}{6}\right)\)

The general solution is

∴ The required solutions are

(ii) 2 cos^{2}θ + 3 sin θ – 3 = θ

Answer:

2 cos^{2}θ + 3 sin θ – 3 = θ

2(1 – sin^{2}θ)+ 3 sin θ – 3 = θ

2 – 2 sin^{2}θ + 3 sin θ – 3 = θ

– 2 sin^{2}θ + 3 sin θ – 1 = θ

2 sin^{2} θ – 3 sin θ + 1 = θ

2 sin^{2}θ – 2 sin θ – sin θ + 1 = θ

2 sin θ (sin θ – 1) – (sin θ – 1) = θ

(2 sin θ – 1) (sin θ – 1) = 0

2 sin θ – 1 = 0 or sin θ – 1 = θ

sin θ = \(\frac { 1 }{ 2 }\) or sin θ = 1

To find the general solution of’ sin θ = \(\frac { 1 }{ 2 }\)

sin θ = \(\frac { 1 }{ 2 }\)

sin θ = sin \(\frac{\pi}{6}\)

The general solution is θ = nπ + (-1)^{n}\(\frac{\pi}{6}\), n ∈ Z

To find the general solution of sin θ = 1

sin θ = 1

sin θ = \(\frac{\pi}{2}\)

The general solution is θ = nπ + (-1)^{n}\(\frac{\pi}{2}\), n ∈ Z

∴ The required solutions are

θ = nπ + (-1)^{n}\(\frac{\pi}{6}\), n ∈ Z (or)

θ = nπ + (-1)^{n}\(\frac{\pi}{6}\), n ∈ Z

(iii) cos θ + cos 3θ = 2 cos 2θ

Answer:

cos 3θ + cos θ = 2 cos 2θ

2 cos 2θ . cos θ = 2 cos 2θ

cos 2θ . cos θ – cos 2θ = θ

cos 2θ (cos θ – 1) = θ

cos 2θ = θ or cos θ – 1 = θ

cos 2θ = θ or cos θ = 1

To find the general solution of cos 2θ = θ

The general solution is

2θ = (2n + 1)\(\frac{\pi}{2}\), n ∈ Z

θ = (2n + 1)\(\frac{\pi}{4}\), n ∈ Z

To find the general solution of cos θ = 1

cos θ = 1

cos θ = cos 0

The general solution is θ = 2nπ , n ∈ Z

∴ The required solutions are

θ = (2n + 1)\(\frac{\pi}{4}\), n ∈ Z (or)

θ = 2nπ, n ∈ Z

(iv) sin θ + sin 3θ + sin 5θ = 0

Answer:

sin 5θ + sin 3θ + sin θ = 0

2 sin 3θ . cos 2θ + sin 3θ = 0

sin 3θ (2 cos 2θ + 1) = θ

sin 3θ = 0 or 2 cos 2θ + 1 = θ

sin 3θ = 0 or cos 2θ = –\(\frac { 1 }{ 2 }\)

To find the general solution of sin 3θ = 0

The general solution is

3θ = nπ, n ∈ Z

θ = \(\frac{\mathbf{n} \pi}{3}\), n ∈ Z

To find the general solution of cos 2θ = –\(\frac { 1 }{ 2 }\)

The general solution is

∴ The required solutions are

(v) sin 2θ – cos 2θ – sin θ + cos θ = θ

Answer:

(vi) sin θ + cos θ = √2

Answer:

The general solution is

(vii) sin θ + √3 cos θ = 1

Answer:

Divide each term by 2

(viii) cot θ + cosec θ = √3

Answer:

(ix) tan θ + tan \(\left(\theta+\frac{\pi}{3}\right)\) + tan \(\left(\theta+\frac{2 \pi}{3}\right)\) = √3

Answer:

(x) cos 2θ = \(\frac{\sqrt{5}+1}{4}\)

Answer:

we know cos 36° = \(\frac{\sqrt{5}+1}{4}\), 36° = \(\frac{\pi}{5}\)

cos 2θ = cos 36° = cos \(\left(\frac{\pi}{5}\right)\)

The general solution is

2θ = 2nπ ± \(\frac{\pi}{5}\), n ∈ Z

θ = nπ ± \(\frac{\pi}{10}\), n ∈ Z

(xi) 2cos ^{2}x – 7 cos x + 3 = 0

Answer:

2 cos^{2}x – 7 cos x + 3 = 0

2 cos^{2}x – 6 cos x – cos x + 3 = 0

2 cos x (cos x – 3) – 1 (cos x – 3) = 0

(2 cos x – 1) (cos x – 3) = 0

2 cos x – 1 = 0 or cos x – 3 = 0

cos x = \(\frac { 1 }{ 2 }\) or cos x = 3

Since – 1 ≤ cos x ≤ 1 , we have

cos x = 3 is not possible.

∴ cos x = \(\frac { 1 }{ 2 }\)

cos x = cos \(\frac{\pi}{3}\)

The general solution is x = 2nπ ± \(\frac{\pi}{3}\), n ∈ Z