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 \leq 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 \leq 2 π

Solve by substitution

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

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

cos (x) = 0, 0 \leq x \leq 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}

tan (θ) = 12

tan (θ) = 15

tan (θ) = 18

arcsin (x) = 2

- 4 cos (2 θ) + 10 sin (θ) + 14 = 7

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