What is the general solution for cos(2x-pi/5)=0 ?

The general solution for cos(2x-pi/5)=0 is x=pin+(7pi)/(20),x=pin+(17pi)/(20)

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cos(2x-pi/5)=0

cos (2 x - \frac{π}{5}) = 0

x = πn + \frac{7 π}{20}, x = πn + \frac{17 π}{20}

Solution steps

cos (2 x - \frac{π}{5}) = 0

General solutions for

cos (2 x - \frac{π}{5}) = 0

2 x - \frac{π}{5} = \frac{π}{2} + 2 πn, 2 x - \frac{π}{5} = \frac{3 π}{2} + 2 πn

Solve

2 x - \frac{π}{5} = \frac{π}{2} + 2 πn : x = πn + \frac{7 π}{20}

Solve

2 x - \frac{π}{5} = \frac{3 π}{2} + 2 πn : x = πn + \frac{17 π}{20}

x = πn + \frac{7 π}{20}, x = πn + \frac{17 π}{20}

cos (a) = sin (a + 3 0^{\circ})

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

\frac{sin ( 5 0 ^{\circ} )}{10} = \frac{sin ( q )}{12}

271.63 = \frac{3 3 \cdot 169.7}{π} cos (α)

2 sin (t) cos (t) + sin (t) - 2 cos (t) - 1 = 0

What is the general solution for cos(2x-pi/5)=0 ?
The general solution for cos(2x-pi/5)=0 is x=pin+(7pi)/(20),x=pin+(17pi)/(20)