What is the general solution for tan(2x-1)=0 ?

The general solution for tan(2x-1)=0 is x=(pin)/2+1/2

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tan(2x-1)=0

tan (2 x - 1) = 0

x = \frac{πn}{2} + \frac{1}{2}

Degrees

x = 28.64788 \dots^{\circ} + 9 0^{\circ} n

Solution steps

tan (2 x - 1) = 0

General solutions for

tan (2 x - 1) = 0

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

2 x - 1 = 0 + πn

2 x - 1 = 0 + πn

Solve

2 x - 1 = 0 + πn : x = \frac{πn}{2} + \frac{1}{2}

2 x - 1 = 0 + πn

0 + πn = πn

2 x - 1 = πn

Move

1

to the right side

2 x - 1 = πn

Add

1

to both sides

2 x - 1 + 1 = πn + 1

Simplify

2 x = πn + 1

2 x = πn + 1

Divide both sides by

2

2 x = πn + 1

Divide both sides by

2

\frac{2 x}{2} = \frac{πn}{2} + \frac{1}{2}

Simplify

x = \frac{πn}{2} + \frac{1}{2}

x = \frac{πn}{2} + \frac{1}{2}

x = \frac{πn}{2} + \frac{1}{2}

cos^{2} (x) = \frac{( 4 k - 3 )}{8}

3 cos^{2} (x) - 8 cos (x) + 4 = 0

sin^{(2)} (x) - sin (x) = 1 - sin^{(2)} (x)

49.55 \cdot 1 - sin^{2} (θ) - 30 sin (θ) = 1.225

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

What is the general solution for tan(2x-1)=0 ?
The general solution for tan(2x-1)=0 is x=(pin)/2+1/2