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= pi/3 ,x=(5pi)/3 ,x=pi

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

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

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

Solution steps

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

Solve by substitution

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

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

cos (x) = - 1, 0 \leq x < 2 π : x = π

Combine all the solutions

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

sin (β) = - 0, 8 (θ \in β v c) s = sec (β) - tan (β)

cos (\frac{2 x - π}{17}) = 0

tan (X) cot (X) - tan (X) + 2 cot (X) = 0

sin (x) = - 0.3926

(sin (2 x) + 1) = 0, 0^{\circ} \leq x < 36 0^{\circ}

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= pi/3 ,x=(5pi)/3 ,x=pi