Chapter 3 Trigonometric Functions Ex-3.1 |
Chapter 3 Trigonometric Functions Ex-3.2 |
Chapter 3 Trigonometric Functions Ex-3.3 |

**Answer
1** :

**Answer
2** :

**Answer
3** :

**Answer
4** :

**Answer
5** :

**Answer
6** :

**Answer
7** :

It is given that

sin 2x + cos x = 0

We can write it as

2 sin x cos x + cos x = 0

cos x (2 sin x + 1) = 0

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

Let cos x = 0

**Answer
8** :

It is given that

sec^{2} 2x= 1 – tan 2x

We can write it as

1 + tan^{2} 2x= 1 – tan 2x

tan^{2} 2x+ tan 2x = 0

Taking common terms

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

Here

tan 2x = 0 or tan 2x +1 = 0

If tan 2x = 0

tan 2x = tan 0

We get

2x = nπ + 0, where n ∈ Z

x = nπ/2, where n ∈ Z

tan 2x + 1 = 0

We can write it as

tan 2x = – 1

So we get

Here

2x = nπ + 3π/4, wheren ∈ Z

x = nπ/2 + 3π/8, wheren ∈ Z

Hence, the generalsolution is nπ/2 or nπ/2 + 3π/8, n ∈ Z.

**Answer
9** :

It is given that

sin x + sin 3x + sin 5x = 0

We can write it as

(sin x + sin 5x) + sin 3x = 0

Using the formula

By further calculation

2 sin 3x cos (-2x) +sin 3x = 0

It can be written as

2 sin 3x cos 2x + sin3x = 0

By taking out thecommon terms

sin 3x (2 cos 2x + 1)= 0

Here

sin 3x = 0 or 2 cos 2x+ 1 = 0

If sin 3x = 0

3x = nπ, where n ∈ Z

We get

x = nπ/3, where n ∈ Z

If 2 cos 2x + 1 = 0

cos 2x = – 1/2

By furthersimplification

= – cos π/3

= cos (π – π/3)

So we get

cos 2x = cos 2π/3

Here

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