**Chapter 3 Trigonometric Functions Ex 3.1**

**Question 1.Find the principal solutions of the following equations :**

We know that,

Hence, the required principal solution are

**(iv) cot θ = 0.Solution:**

**Question 2.Find the principal solutions of the following equations:**(i) sinθ = −1/2

Solution:

We know that,

sin π/6 = 1/2 and sin (π + θ) = -sinθ,

sin(2π – θ) = -sinθ

Hence, the required principal solutions are

θ = 7π/6 and θ = 11π/6.

**(ii) tanθ = -1Solution:**

We know that,

tan π/4 = 1 and tan(π – θ) = -tanθ,

tan(2π – θ) = -tanθ

Hence, the required principal solutions are

θ = 3π/4 and θ = 7π/4.

**(iii) √3 cosecθ + 2 = 0.Solution:**

**Question 3.Find the general solutions of the following equations :**

**Question 4.Find the general solutions of the following equations:**

(ii) cosecθ = –√2

Solution:

The general solution of sinθ = sin∝ is

**(iii) tanθ = -1Solution:**

The general solution of tanθ = tan∝ is

**Question 5.Find the general solutions of the following equations :(i) sin 2θ = 1/2Solution:**

The general solution of sin θ = sin ∝ is

θ = nπ + (-1)

^{n}∝, n ∈ Z

**(ii) tan 2θ/3 = √3Solution:**

The general solution of tan θ = tan ∝ is

**(iii) cot 4θ = -1Solution:**

The general solution of tan θ = tan ∝ is

**Question 6.Find the general solutions of the following equations :(i) 4 cos**

^{2}θ = 3Solution:

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 = 1Solution:**

The general solution of sin

^{2}θ = sin

^{2}∝ is

θ = nπ ± ∝, n ∈ Z

Now, 4 sin

^{2}θ = 3

**(iii) cos 4θ = cos 2θSolution:**

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 = nπ/3, n ∈ Z

Hence, the required general solution is

θ = nπ/3, n ∈ Z or ∴ θ = nπ, n ∈ Z.

Alternative Method:

cos 4θ = cos 2θ

∴ cos4θ – cos 20 = 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 = nπ3, n ∈ Z or θ = nπ, n ∈ Z.

**Question 7.Find the general solutions of the following equations :(i) sinθ = tanθSolution:**sin θ = tan θ

∴ sin θ = sinθ/cosθ

∴ 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 = 3tanθSolution:**

**(iii) cosθ + sinθ = 1.Solution:**

cosθ + sinθ = 1

**Question 8.Which of the following equations have solutions ?(i) cos 2θ = -1Solution:**

cos 2θ = -1

Since -1 ≤ cos θ ≤ 1 for any θ,

cos 2θ = -1 has solution.

**(ii) cos ^{2}θ = -1Solution:**

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 = 3Solution:**

2 sin θ = 3 ∴ sin θ = 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 = 5Solution:**

3tanθ = 5 ∴ tanθ = 5/3

This is possible because tan θ is any real number.

∴ 3tanθ = 5 has solution.