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

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

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

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

x = \frac{2 π}{3}, x = \frac{4 π}{3}

Solution steps

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

Solve by substitution

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

cos (x) = 1, (0, 2 π) :

No Solution

cos (x) = - \frac{1}{2}, (0, 2 π) : x = \frac{2 π}{3}, x = \frac{4 π}{3}

Combine all the solutions

x = \frac{2 π}{3}, x = \frac{4 π}{3}

- 3 + cos (2 x) \cdot 2 = 0

\frac{1}{9} + cos^{2} (x) = 1

sin (x) = 0.9063

11 = 12 + 3.5 sin (\frac{2 π}{365} t)

sin (2 t) = 2 sin (t)

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