Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Ex 3.1 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Ex 3.1

Question 1.

Find the principal solutions of the following equations :

(i) cos θ= \(\frac{1}{2}\)

Solution:

We know that, cos\(\frac{\pi}{3}\) = \(\frac{1}{2}\) and cos (2π – θ) = cos θ

∴ cos\(\frac{\pi}{3}\) = cos(2π – \(\frac{\pi}{3}\)) = cos\(\frac{5 \pi}{3}\)

∴ cos\(\frac{\pi}{3}\) = cos\(\frac{5 \pi}{3}\) = \(\frac{1}{2}\), where

0 < \(\frac{\pi}{3}\) < 2π and 0 < \(\frac{5 \pi}{3}\) < 2π

∴ cos θ = \(\frac{1}{2}\) gives cos θ = cos\(\frac{\pi}{3}\) = cos\(\frac{5 \pi}{3}\)

∴ θ = \(\frac{\pi}{3}\) and θ = \(\frac{5 \pi}{3}\)

Hence, the required principal solutions are

θ = \(\frac{\pi}{3}\) and θ = \(\frac{5 \pi}{3}\)

(ii) sec θ = \(\frac{2}{\sqrt{3}}\)

Solution:

(iii) cot θ = \(\sqrt {3}\)

Solution:

The given equation is cot θ = \(\sqrt {3}\) which is same as tan θ = \(\frac{1}{\sqrt{3}}\).

We know that,

Hence, the required principal solution are

θ = \(\frac{\pi}{6}\) and θ = \(\frac{7 \pi}{6}\).

(iv) cot θ = 0.

Solution:

Question 2.

Find the principal solutions of the following equations:

(i) sinθ = \(-\frac{1}{2}\)

Solution:

We know that,

sin\(\frac{\pi}{6}\) = \(\frac{1}{2}\) and sin (π + θ) = -sinθ,

sin(2π – θ) = -sinθ

Hence, the required principal solutions are

θ = \(\frac{7\pi}{6}\) and θ = \(\frac{11 \pi}{6}\).

(ii) tanθ = -1

Solution:

We know that,

tan\(\frac{\pi}{4}\) = 1 and tan(π – θ) = -tanθ,

tan(2π – θ) = -tanθ

Hence, the required principal solutions are

θ = \(\frac{3\pi}{4}\) and θ = \(\frac{7 \pi}{4}\).

(iii) \(\sqrt {3}\) cosecθ + 2 = 0.

Solution:

Question 3.

Find the general solutions of the following equations :

(i) sinθ = \(\frac{1}{2}\)

Solution:

(i) The general solution of sin θ = sin ∝ is

θ = nπ + (-1 )^{n}∝, n ∈ Z

Now, sinθ = \(\frac{1}{2}\) = sin\(\frac{\pi}{6}\) …[∵ sin\(\frac{\pi}{6}\) = \(\frac{1}{2}\)]

∴ the required general solution is

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

(ii) cosθ = \(\frac{\sqrt{3}}{2}\)

Solution:

The general solution of cos θ = cos ∝ is

θ = 2nπ ± ∝, n ∈ Z

Now, cosθ = \(\frac{\sqrt{3}}{2}\) = cos\(\frac{\pi}{6}\) …[∵ cos\(\frac{\pi}{6}\) = \(\frac{\sqrt{3}}{2}\)]

∴ the required general solution is

θ = 2nπ ± \(\frac{\pi}{6}\), n ∈ Z.

(iii) tanθ = \(\frac{1}{\sqrt{3}}\)

Solution:

The general solution of tan θ = tan ∝ is

θ = nπ + ∝, n ∈ Z

Now, tan θ = \(\frac{1}{\sqrt{3}}\) = tan\(\frac{\pi}{6}\) …[tan\(\frac{\pi}{6}\) = \(\frac{1}{\sqrt{3}}\)]

∴ the required general solution is

θ = nπ + \(\frac{\pi}{6}\) , n ∈ Z.

(iv) cotθ = 0.

Solution:

The general solution of tan θ = tan ∝ is

θ = nπ + ∝, n ∈ Z

Now, cot θ = 0 ∴ tan θ does not exist

∴ tanθ = tan\(\frac{\pi}{2}\) [∵ tan\(\frac{\pi}{2}\) does not exist]

∴ the required general solution is

θ = nπ + \(\frac{\pi}{2}\), n ∈ Z.

Question 4.

Find the general solutions of the following equations:

(i) secθ = \(\sqrt {2}\)

Solution:

The general solution of cos θ = cos ∝ is

θ = nπ ± ∝, n ∈ Z.

Now, secθ = \(\sqrt {2}\) ∴ cosθ = \(\frac{1}{\sqrt{2}}\)

∴ cosθ = cos\(\frac{\pi}{4}\) ….[cos\(\frac{\pi}{4}\) = \(\frac{1}{\sqrt{2}}\)]

∴ the required general solution is

θ = 2nπ ± \(\frac{\pi}{4}\), n ∈ Z.

(ii) cosecθ = –\(\sqrt {2}\)

Solution:

The general solution of sinθ = sin∝ is

(iii) tanθ = -1

Solution:

The general solution of tanθ = tan∝ is

Question 5.

Find the general solutions of the following equations :

(i) sin 2θ = \(\frac{1}{2}\)

Solution:

The general solution of sin θ = sin ∝ is

θ = nπ + (-1)^{n}∝, n ∈ Z

(ii) tan \(\frac{2 \theta}{3}\) = \(\sqrt {3}\)

Solution:

The general solution of tan θ = tan ∝ is

(iii) cot 4θ = -1

Solution:

The general solution of tan θ = tan ∝ is

Question 6.

Find the general solutions of the following equations :

(i) 4 cos^{2}θ = 3

Solution:

The general solution of cos^{2}θ = cos^{2} ∝ is

θ = nπ ± ∝, n ∈ Z

Now, 4 cos^{2}θ = 3

(ii) 4 sin^{2}θ = 1

Solution:

The general solution of sin^{2}θ = sin^{2} ∝ is

θ = nπ ± ∝, n ∈ Z

Now, 4 sin^{2}θ = 3

(iii) cos 4θ = cos 2θ

Solution:

The general solution of cos θ = cos ∝ is

θ = 2nπ ± ∝, n ∈ Z

∴ the general solution of cos 4θ = cos 2θ is given by

4θ = 2nπ ± 2θ, n ∈ Z

Taking positive sign, we get

4θ = 2nπ + 2θ, n ∈ Z

∴ 2θ = 2nπ, n ∈ Z

∴ θ = nπ, n ∈ Z

Taking negative sign, we get

4θ = 2nπ – 2θ, n ∈ Z

∴ 6θ = 2nπ, n ∈ Z

∴ θ = \(\frac{n \pi}{3}\), n ∈ Z

Hence, the required general solution is

θ = \(\frac{n \pi}{3}\), n ∈ Z or ∴ θ = nπ, n ∈ Z.

Alternative Method:

cos 4θ = cos 2θ

∴ cos4θ – cos 20 = 0

∴ -2sin\(\left(\frac{4 \theta+2 \theta}{2}\right)\)∙sin\(\left(\frac{4 \theta-2 \theta}{2}\right)\) = 0

∴ sin3θ∙sinθ = 0

∴ either sin3θ = 0 or sin θ = 0

The general solution of sin θ = 0 is

θ = nπ, n ∈ Z.

∴ the required general solution is given by

3θ = nπ, n ∈ Z or θ = nπ, n ∈ Z

i.e. θ = \(\frac{n \pi}{3}\), n ∈ Z or θ = nπ, n ∈ Z.

Question 7.

Find the general solutions of the following equations :

(i) sinθ = tanθ

Solution:

sin θ = tan θ

∴ sin θ = \(\frac{\sin \theta}{\cos \theta}\)

∴ sin θ cos θ = sin θ

∴ sin θ cos θ – sinθ = 0

∴ sin θ (cos θ – 1) = θ

∴ either sinθ = 0 or cosθ – 1 = 0

∴ either sin θ = 0 or cos θ = 1

∴ either sinθ = 0 or cosθ = cosθ …[∵ cos0 = 1]

The general solution of sinθ = 0 is θ = nπ, n ∈ Z and cos θ = cos ∝ is θ = 2nπ ± ∝, where n ∈ Z.

∴ the required general solution is given by

θ = nπ, n ∈ Z or θ = 2nπ ± 0, n ∈ Z

∴ θ = nπ, n ∈ Z or θ = 2nπ, n ∈ Z.

(ii) tan^{3}θ = 3tanθ

Solution:

tan^{3}θ = 3tanθ

∴ tan^{3}θ – 3tanθ = 0

∴ tan θ (tan^{2}θ – 3) = 0

∴ either tan θ = 0 or tan^{2}θ – 3 = 0

∴ either tanθ = 0 or tan^{2}θ = 3

∴ either tan θ = 0 or tan^{2}θ = (\(\sqrt {3}\) )^{3}

∴ either tan θ = 0 or tan^{2}θ = (tan\(\frac{\pi}{3}\))^{3} …[tan\(\frac{\pi}{3}\) = \(\sqrt {3}\)]

∴ either tanθ = 0 or tan^{2}θ = tan^{2}\(\frac{\pi}{3}\)

The general solution of

tanθ = 0 is θ = nπ, n ∈ Z and

tan^{2}θ = tan^{2}∝ is θ = nπ ± ∝, n ∈ Z.

∴ the required general solution is given by

θ = nπ, n ∈ Z or θ = nπ ± \(\frac{\pi}{3}\), n ∈ Z.

(iii) cosθ + sinθ = 1.

Solution:

cosθ + sinθ = 1

Question 8.

Which of the following equations have solutions ?

(i) cos 2θ = -1

Solution:

cos 2θ = -1

Since -1 ≤ cos θ ≤ 1 for any θ,

cos 2θ = -1 has solution.

(ii) cos^{2}θ = -1

Solution:

cos^{2}θ = -1

This is not possible because cos^{2}θ ≥ 0 for any θ.

∴ cos^{2}θ = -1 does not have any solution.

(iii) 2 sinθ = 3

Solution:

2 sin θ = 3 ∴ sin θ = \(\frac{3}{2}\)

This is not possible because -1 ≤ sin θ ≤ 1 for any θ.

∴ 2 sin θ = 3 does not have any solution.

(iv) 3 tanθ = 5

Solution:

3tanθ = 5 ∴ tanθ = \(\frac{5}{3}\)

This is possible because tan θ is any real number.

∴ 3tanθ = 5 has solution.