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= pi/3 ,x=pi,x=(5pi)/3

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

Solution steps

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

Rewrite using trig identities

2 cos (\frac{3 x}{2}) cos (\frac{x}{2}) = 0

Solving each part separately

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

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

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

Combine all the solutions

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

cos (5 x) cos (3 x) - sin (5 x) sin (3 x) = \frac{3}{2}

- 6 sin (2 x) = 0

sin (3 x) - 1 = 0

- sin (θ) = 0

sin (θ) = \frac{1}{6}

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= pi/3 ,x=pi,x=(5pi)/3