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

The general solution for 2cos^2(x)-cos(x)-1=0,0<= x<= 2pi is x=0,x=2pi,x=(2pi)/3 ,x=(4pi)/3

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

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

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

Solution steps

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

Solve by substitution

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

cos (x) = 1, 0 \leq x \leq 2 π : x = 0, x = 2 π

cos (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{2 π}{3}, x = \frac{4 π}{3}

Combine all the solutions

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

sin (θ) = - \frac{6}{7}

a sin (θ) = b cos (θ)

- 9 tan^{2} (θ) + 12 tan (θ) - 4 = 0

- 5 cos (x) = 0

2 sin^{2} (θ) - sin (θ) = 3

What is the general solution for 2cos^2(x)-cos(x)-1=0,0<= x<= 2pi ?
The general solution for 2cos^2(x)-cos(x)-1=0,0<= x<= 2pi is x=0,x=2pi,x=(2pi)/3 ,x=(4pi)/3