What is the general solution for sin(2x)-sin(x)=0,0<= x<= 2pi ?

The general solution for sin(2x)-sin(x)=0,0<= x<= 2pi is x=0,x=pi,x=2pi,x= pi/3 ,x=(5pi)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(2x)-sin(x)=0,0<= x<= 2pi

sin (2 x) - sin (x) = 0, 0 \leq x \leq 2 π

x = 0, x = π, x = 2 π, x = \frac{π}{3}, x = \frac{5 π}{3}

Solution steps

sin (2 x) - sin (x) = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

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

Factor

- sin (x) + 2 cos (x) sin (x) : sin (x) (2 cos (x) - 1)

sin (x) (2 cos (x) - 1) = 0

Solving each part separately

sin (x) = 0 or 2 cos (x) - 1 = 0

sin (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

2 cos (x) - 1 = 0, 0 \leq x \leq 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

Combine all the solutions

x = 0, x = π, x = 2 π, x = \frac{π}{3}, x = \frac{5 π}{3}

- 2 cos (3 θ) = 1

sec^{2} (x) + 0.5 tan (x) - 1 = 0

cos (2 x) = sin (x), 0 \leq x \leq 2 π

tan (x - \frac{π}{2}) = 1

5 cos (x) = 4 sin (x) + 4

What is the general solution for sin(2x)-sin(x)=0,0<= x<= 2pi ?
The general solution for sin(2x)-sin(x)=0,0<= x<= 2pi is x=0,x=pi,x=2pi,x= pi/3 ,x=(5pi)/3