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

The general solution for -sqrt(2)tan(θ)-sec(θ)=0 is θ=(5pi)/4+2pin,θ=(7pi)/4+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-sqrt(2)tan(θ)-sec(θ)=0

Solution

- 2 tan (θ) - sec (θ) = 0

Solution

θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Degrees

θ = 22 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\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 \frac{sin ( θ )}{cos ( θ )}

Simplify

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

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

Multiply

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

2 \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

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

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

- 1 - sin (θ) 2 = 0

Move

1

to the right side

- 1 - sin (θ) 2 = 0

Add

1

to both sides

- 1 - sin (θ) 2 + 1 = 0 + 1

Simplify

- sin (θ) 2 = 1

- sin (θ) 2 = 1

Divide both sides by

- 2

- sin (θ) 2 = 1

Divide both sides by

- 2

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

Simplify

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

Simplify

\frac{- sin ( θ ) 2}{- 2} : sin (θ)

\frac{- sin ( θ ) 2}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{sin ( θ ) 2}{2}

Cancel the common factor:

2

= sin (θ)

Simplify

\frac{1}{- 2} : - \frac{2}{2}

\frac{1}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{1}{2}

Rationalize

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

sin (θ) = - \frac{2}{2}

sin (θ) = - \frac{2}{2}

sin (θ) = - \frac{2}{2}

General solutions for

sin (θ) = - \frac{2}{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{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Popular Examples

4cos(x)=-sin^2(x)+1,0<= x<= 2pi

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

2*sin(x)=tan(x)

2 \cdot sin (x) = tan (x)

0=arcsin(x)

0 = arcsin (x)

cos(x)=0.5149,0<= x<2pi

cos (x) = 0.5149, 0 \leq x < 2 π

sin(θ)-0.2cos(θ)=0.704

sin (θ) - 0.2 cos (θ) = 0.704

Frequently Asked Questions (FAQ)

What is the general solution for -sqrt(2)tan(θ)-sec(θ)=0 ?
The general solution for -sqrt(2)tan(θ)-sec(θ)=0 is θ=(5pi)/4+2pin,θ=(7pi)/4+2pin