What is the general solution for 2tan((pix)/2)=sec((pix)/2) ?

The general solution for 2tan((pix)/2)=sec((pix)/2) is x=(1+12n)/3 ,x=(5+12n)/3

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2tan((pix)/2)=sec((pix)/2)

2 tan (\frac{π x}{2}) = sec (\frac{π x}{2})

x = \frac{1 + 12 n}{3}, x = \frac{5 + 12 n}{3}

Solution steps

2 tan (\frac{π x}{2}) = sec (\frac{π x}{2})

Subtract

sec (\frac{π x}{2})

from both sides

2 tan (\frac{π x}{2}) - sec (\frac{π x}{2}) = 0

Let:

u = \frac{π x}{2}

2 tan (u) - sec (u) = 0

Express with sin, cos

\frac{- 1 + 2 sin ( u )}{cos ( u )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 1 + 2 sin (u) = 0

Move

1

to the right side

2 sin (u) = 1

Divide both sides by

2

sin (u) = \frac{1}{2}

General solutions for

sin (u) = \frac{1}{2}

u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

Substitute back

u = \frac{π x}{2}

\frac{π x}{2} = \frac{π}{6} + 2 πn : x = \frac{1 + 12 n}{3}

\frac{π x}{2} = \frac{5 π}{6} + 2 πn : x = \frac{5 + 12 n}{3}

x = \frac{1 + 12 n}{3}, x = \frac{5 + 12 n}{3}

1.47 = (\frac{sin ( \frac{x + 60}{2} )}{sin ( 3 0 ^{\circ} )})

sin^{2} (x) + 1 = 3 sin (x) - 1

2 tan (θ) = sec^{2} (θ)

tan (θ) = \frac{14}{15}

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

What is the general solution for 2tan((pix)/2)=sec((pix)/2) ?
The general solution for 2tan((pix)/2)=sec((pix)/2) is x=(1+12n)/3 ,x=(5+12n)/3