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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

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

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

Combine all the solutions

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

x, arctan (x) = 4 t + c

cos (\frac{π}{3} - x) = 1

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

tan (x) = \frac{6}{15}

2 cos^{3} (x) + cos^{2} (x) - 2 cos (x) - 1 = 0

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