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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Degrees

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

Solution steps

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

Factor

2 cos (x) - cos (x) sin (x) : - cos (x) (sin (x) - 2)

2 cos (x) - cos (x) sin (x)

Factor out common term

- cos (x)

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

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

Solving each part separately

cos (x) = 0 or sin (x) - 2 = 0

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

cos (x) = 0

General solutions for

cos (x) = 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}

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

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

sin (x) - 2 = 0 :

No Solution

sin (x) - 2 = 0

Move

2

to the right side

sin (x) - 2 = 0

Add

2

to both sides

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

Simplify

sin (x) = 2

sin (x) = 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

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

10 sin (x) - 7 = 0

2 tan^{3} (x) - 6 tan (x) = 0, 0 \leq x \leq π

5 cos (x) = 5 cos (2 x)

cos (4 θ) = cos (2 θ)

3 cos (3 x) = 2

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