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/2 ,x=(3pi)/2 ,x=(2pi)/3 ,x=(4pi)/3

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

Solution steps

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

Solve by substitution

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

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

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

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

2 cos (θ) - sin (θ) = 0

cos (x + \frac{π}{6}) = sin (x)

sin (θ) = - \frac{1}{5}

5 cot (x) - 5 = 0

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/2 ,x=(3pi)/2 ,x=(2pi)/3 ,x=(4pi)/3