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 < 2 π

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

Solution steps

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

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

Combine all the solutions

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

10 cos^{2} (x) = 5

cos (θ) = \frac{5}{9}

sin^{2} (x) = 5 (cos (x) + 1)

tan (2 x - \frac{π}{6}) = - \frac{1}{3}

4 cos^{2} (2 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= pi/3 ,x=pi,x=(5pi)/3