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

The general solution for 2cos^2(x)+cos(x)-1=0,0<= x<2pi is x= pi/3 ,x=(5pi)/3 ,x=pi

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

2 cos^{2} (x) + cos (x) - 1 = 0, 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, 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

cos (π x) = 0

cos^{3} (0)

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

cos (24 0^{\circ})

arcsin (sin (\frac{7 π}{6}))

What is the general solution for 2cos^2(x)+cos(x)-1=0,0<= x<2pi ?
The general solution for 2cos^2(x)+cos(x)-1=0,0<= x<2pi is x= pi/3 ,x=(5pi)/3 ,x=pi