What is the general solution for 2tan(θ)-sec(θ)=0 ?

The general solution for 2tan(θ)-sec(θ)=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin

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2tan(θ)-sec(θ)=0

2 tan (θ) - sec (θ) = 0

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

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

2 tan (θ) - sec (θ) = 0

Express with sin, cos

- sec (θ) + 2 tan (θ)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

= - \frac{1}{cos ( θ )} + 2 tan (θ)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - \frac{1}{cos ( θ )} + 2 \cdot \frac{sin ( θ )}{cos ( θ )}

Simplify

- \frac{1}{cos ( θ )} + 2 \cdot \frac{sin ( θ )}{cos ( θ )} : \frac{- 1 + 2 sin ( θ )}{cos ( θ )}

- \frac{1}{cos ( θ )} + 2 \cdot \frac{sin ( θ )}{cos ( θ )}

Multiply

2 \cdot \frac{sin ( θ )}{cos ( θ )} : \frac{2 sin ( θ )}{cos ( θ )}

2 \cdot \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{sin ( θ ) \cdot 2}{cos ( θ )}

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

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 + 2 sin ( θ )}{cos ( θ )}

= \frac{- 1 + 2 sin ( θ )}{cos ( θ )}

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

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

- 1 + 2 sin (θ) = 0

Move

1

to the right side

- 1 + 2 sin (θ) = 0

Add

1

to both sides

- 1 + 2 sin (θ) + 1 = 0 + 1

Simplify

2 sin (θ) = 1

2 sin (θ) = 1

Divide both sides by

2

2 sin (θ) = 1

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{1}{2}

Simplify

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \frac{1}{2}

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

sin (θ) = \frac{175}{214}

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

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

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

arccosh (x) = \frac{2 π}{7}

What is the general solution for 2tan(θ)-sec(θ)=0 ?
The general solution for 2tan(θ)-sec(θ)=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin