What is the general solution for 2*cos^2(x)-cos(x)-1=0,0<= x<= pi ?

The general solution for 2*cos^2(x)-cos(x)-1=0,0<= x<= pi is x=0,x=(2pi)/3

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2*cos^2(x)-cos(x)-1=0,0<= x<= pi

2 \cdot cos^{2} (x) - cos (x) - 1 = 0, 0 \leq x \leq π

x = 0, x = \frac{2 π}{3}

Solution steps

2 cos^{2} (x) - cos (x) - 1 = 0, 0 \leq x \leq π

Solve by substitution

cos (x) = 1, cos (x) = - \frac{1}{2}

cos (x) = 1, 0 \leq x \leq π : x = 0

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

Combine all the solutions

x = 0, x = \frac{2 π}{3}

2 cos^{3} (x) - cos^{2} (x) = 0

sin^{2} (x) - 2 = sin (x)

arccosh (x) = \frac{2 π}{7}

arccosh (x) = \frac{2 π}{5}

- 6 sin (x) - 4 sin (2 x) = 0

What is the general solution for 2*cos^2(x)-cos(x)-1=0,0<= x<= pi ?
The general solution for 2*cos^2(x)-cos(x)-1=0,0<= x<= pi is x=0,x=(2pi)/3