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

The general solution for cos(2x)-cos(x)=0,0<= x<= 2pi is x=0,x=(2pi)/3 ,x=(4pi)/3 ,x=2pi

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

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

x = 0, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 2 π

Solution steps

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

Rewrite using trig identities

- 2 sin (\frac{3 x}{2}) sin (\frac{x}{2}) = 0

Solving each part separately

sin (\frac{3 x}{2}) = 0 or sin (\frac{x}{2}) = 0

sin (\frac{3 x}{2}) = 0, 0 \leq x \leq 2 π : x = 0, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 2 π

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

Combine all the solutions

x = 0, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 2 π

cos (θ) = 2

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

cos (4 x) - 1 = 0

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

cos^{2} (x) - cos (x) - 2 = 0

What is the general solution for cos(2x)-cos(x)=0,0<= x<= 2pi ?
The general solution for cos(2x)-cos(x)=0,0<= x<= 2pi is x=0,x=(2pi)/3 ,x=(4pi)/3 ,x=2pi