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

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

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

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

x = \frac{π}{3}, x = \frac{5 π}{3}, x = \frac{π}{2}, x = \frac{3 π}{2}

Solution steps

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

Solve by substitution

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

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

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

Combine all the solutions

x = \frac{π}{3}, x = \frac{5 π}{3}, x = \frac{π}{2}, x = \frac{3 π}{2}

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

14 sin^{2} (x) = 7

csc (x) = sin (x)

tan (2 x + \frac{π}{3}) = 0

18 cos^{2} (x) = 9

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