What is the general solution for cos(θ)sin(θ)-2cos(θ)=0 ?

The general solution for cos(θ)sin(θ)-2cos(θ)=0 is θ= pi/2+2pin,θ=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(θ)sin(θ)-2cos(θ)=0

cos (θ) sin (θ) - 2 cos (θ) = 0

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (θ) sin (θ) - 2 cos (θ) = 0

Factor

cos (θ) sin (θ) - 2 cos (θ) : cos (θ) (sin (θ) - 2)

cos (θ) sin (θ) - 2 cos (θ)

Factor out common term

cos (θ)

= cos (θ) (sin (θ) - 2)

cos (θ) (sin (θ) - 2) = 0

Solving each part separately

cos (θ) = 0 or sin (θ) - 2 = 0

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

sin (θ) - 2 = 0 :

No Solution

sin (θ) - 2 = 0

Move

2

to the right side

sin (θ) - 2 = 0

Add

2

to both sides

sin (θ) - 2 + 2 = 0 + 2

Simplify

sin (θ) = 2

sin (θ) = 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

y, x = tan^{2} (y)

tan (θ) = \frac{7}{3}

sin (4 θ) - sin (2 θ) = 0

tan (x) sin (x) - tan (x) = 0

cos^{2} (θ) - 7 cos (θ) + 12 = 0

What is the general solution for cos(θ)sin(θ)-2cos(θ)=0 ?
The general solution for cos(θ)sin(θ)-2cos(θ)=0 is θ= pi/2+2pin,θ=(3pi)/2+2pin