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 \leq 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 \leq 2 π

Solve by substitution

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

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

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

Combine all the solutions

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

arcsin (x) - arccos (- \frac{1}{2}) = \frac{π}{6}

tan^{2} (x) - tan (x) - 2 = 0

tan^{3} (x) = 3 tan (x)

sin (2 x) = - cos (2 x)

3 sin (4 x) + 5 = 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