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

The general solution for 2sin^2(x)-cos(x)-1=0,0<= x<= 2pi is x=pi,x= pi/3 ,x=(5pi)/3

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

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

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

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

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

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

Combine all the solutions

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

5 tan (x) sin (x) = - sin (x)

2 sin (4 x) - 1 = 0

cot^{2} (x) - cot (x) = 0

2 cos (\frac{x}{2}) + 1 = 0

- tan (2 θ) + 2 sin (θ) = 0

What is the general solution for 2sin^2(x)-cos(x)-1=0,0<= x<= 2pi ?
The general solution for 2sin^2(x)-cos(x)-1=0,0<= x<= 2pi is x=pi,x= pi/3 ,x=(5pi)/3