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

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

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

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

sin (t) + 2 sin (2 t) = 0

1 + cos (θ) = 0

x, (D^{2} - 3 D + 2) y = cos (e^{- x})

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

cos^{2} (θ) - \frac{1}{4} = 0

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