### Find the principal and general solutions of the following equations:

### Question 1. tan x = √3

**Solution:**

Given:tan x = √3Here, x lies in first or third quadrant.

∴ tan

x= tan60°∘ or tanx= tan(180° + 60°)tan

x= tan60°∘ or tanx= tan240°∘Here, the principal solutions are

π/3, 4π/3.∴ tan

x= tanπ/3

⇒ x = nπ + π/3 where n ∈ Z

### Question 2. sec x = 2

**Solution:**

Given:sec x = 2It can be written as, cos x = 1/2

Here, x lies in first or fourth quadrant.

∴ cos

x= cos60°∘ or cosx= cos(360° – 60°)cos

x= cos60°∘ or cosx= cos300°∘cosx = cos

π/3 or cosx = cos 5π/3Here, the principal solutions are π/3, 5 π/3

∴ cosx = cosπ

x= 2nπ±π/3 where,n∈Z

### Question 3. cot x = −√3

**Solution:**

Given:cot x = −√3It can be written as, tan x = -1/√3

Here, x lies in second or fourth quadrant.

∴ tanx = −tan30°∘= tan(180°– 30°) or tanx = tan(360°∘- 60°)

tanx = tan150°∘ or tanx = tan330°∘

tanx = tan5π/6 or tanx = tan11π/6Here, the principal solutions are 5π/6, 11 π/6

∴ tan x = tan 5π/6

⇒

x=nπ+ 5π/3 where,n∈Z

### Question 4. cosec x = – 2

**Solution:**

Given:cosec x = – 2It can be written as, sin x = -1/2

Here x lies in third or fourth quadrant.

∴ sinx = -sin30°∘= sin(180°+ 30°) or sinx = sin(360°∘- 30°)

sinx = sin210°∘ or sinx = sin330°∘

sinx = sin7π/6 or sinx = sin11π/6Here, the principal solutions are 7π/6, 11 π/6

∴ sin x = -sin π/6

⇒

x=nπ+ (−1)7^{n}π/6 where,n∈Z

### Find the general solution for each of the following equations:

### Question 5. cos 4x = cos 2x

**Solution:**

Given:cos 4x = cos 2x4

x= 2nπ± 2x,n∈Z4

x –2x= 2nπor 4x+ 2x= 2nπ,n∈Z2

x= 2nπor 6x= 2nπ,n∈Z2

x= 2nπor 6x= 2nπ,n∈Z

x=nπorx= 3nπ,n∈Z

Therefore,theprincipalsolutionsarenπ,nπ/3

### Question 6. cos 3x + cos x – cos 2x = 0

**Solution:**

Given:cos 3x + cos x – cos 2x = 02

cos2xcosx – cos2x= 0

cos2x(2cosx –1) = 0

cos2x= 0 or 2cosx –1 = 02

x= (2n+ 1)π/2 orcosx= 1/2 =cosπ/3,n∈z

x= (2n+ 1)π/4 orx= 2nπ± 2π/3,n∈z

### Question 7. sin 2x + cos x = 0

**Solution:**

Given:sin 2x + cos x = 02

sinxcosx+cosx= 0

cosx(2sinx+ 1) = 0

cosx= 0 or 2sinx+ 1 = 0

x= (2n+ 1)π/2 orsinx= −1/2 = −sinπ/6,n∈z

x= (2n+ 1)π/4 orx= 2nπ± 2π/3,n∈z

x= (2n+ 1)π/4 orx=nπ+ (-1),^{n}-π/6n∈Z

x= (2n+ 1)π/4 orx=nπ+ (-1)7^{n}π/6,n∈Z

### Question 8. sec^{2}2x = 1 – tan 2x

**Solution:**

Given:sec^{2}2x = 1 – tan 2x1 +

tan^{2}2x= 1 –tan2x

tan2x(tan2x+ 1) = 0

tan2x= 0 ortan2x+ 1 = 02

x=nπortan2x –1 = –tanπ/4

x=nπ/2 orx=nπ/2 + 3π/8,n=Z

### Question 9. sin x + sin 3x + sin 5x = 0

**Solution:**

Given:sin x + sin 3x + sin 5x = 02

sin3xcos2x+sin3x= 0

sin3x(2cos2x+ 1) = 0

sin3x= 0 or 2cos2x+ 1 = 13

x=nπorcos2x= -1/2 =cos2π/3,n∈z

x=nπ/3 or 2x=nπ/2± 2π/3,n∈z

x=nπ/3 orx=nπ±π/3,n∈z