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

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

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

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

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

Solution steps

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

Solve by substitution

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

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

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

7 sin^{2} (θ) - 36 sin (θ) + 5 = 0

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

tan (\frac{2 x}{3}) = 0

3 cos^{2} (A) + 3 cos (A) = 0

3 csc (2 x) - 2 = 0

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