What is the general solution for tan(θ)+sec(θ)=2cos(θ) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan(θ)+sec(θ)=2cos(θ)

Solution

tan (θ) + sec (θ) = 2 cos (θ)

Solution

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

tan (θ) + sec (θ) = 2 cos (θ)

Subtract

2 cos (θ)

from both sides

tan (θ) + sec (θ) - 2 cos (θ) = 0

Express with sin, cos

sec (θ) + tan (θ) - 2 cos (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Combine the fractions

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

Apply rule

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

= \frac{1 + sin ( θ )}{cos ( θ )}

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

Convert element to fraction:

2 cos (θ) = \frac{2 cos ( θ ) cos ( θ )}{cos ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

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

1 + sin (θ) - 2 cos (θ) cos (θ)

2 cos (θ) cos (θ) = 2 cos^{2} (θ)

2 cos (θ) cos (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= 2 cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 2 cos^{2} (θ)

= 1 + sin (θ) - 2 cos^{2} (θ)

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

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

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

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

1 + sin (θ) - 2 cos^{2} (θ) = 0

Rewrite using trig identities

1 + sin (θ) - 2 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= 1 + sin (θ) - 2 (1 - sin^{2} (θ))

Simplify

1 + sin (θ) - 2 (1 - sin^{2} (θ)) : 2 sin^{2} (θ) + sin (θ) - 1

1 + sin (θ) - 2 (1 - sin^{2} (θ))

Expand

- 2 (1 - sin^{2} (θ)) : - 2 + 2 sin^{2} (θ)

- 2 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 1, c = sin^{2} (θ)

= - 2 \cdot 1 - (- 2) sin^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 + 2 sin^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 sin^{2} (θ)

= 1 + sin (θ) - 2 + 2 sin^{2} (θ)

Simplify

1 + sin (θ) - 2 + 2 sin^{2} (θ) : 2 sin^{2} (θ) + sin (θ) - 1

1 + sin (θ) - 2 + 2 sin^{2} (θ)

Group like terms

= sin (θ) + 2 sin^{2} (θ) + 1 - 2

Add/Subtract the numbers:

1 - 2 = - 1

= 2 sin^{2} (θ) + sin (θ) - 1

= 2 sin^{2} (θ) + sin (θ) - 1

= 2 sin^{2} (θ) + sin (θ) - 1

- 1 + sin (θ) + 2 sin^{2} (θ) = 0

Solve by substitution

- 1 + sin (θ) + 2 sin^{2} (θ) = 0

Let:

sin (θ) = u

- 1 + u + 2 u^{2} = 0

- 1 + u + 2 u^{2} = 0 : u = \frac{1}{2}, u = - 1

- 1 + u + 2 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

2 u^{2} + u - 1 = 0

Solve with the quadratic formula

2 u^{2} + u - 1 = 0

Quadratic Equation Formula:

For

a = 2, b = 1, c = - 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 1) = 3

1^{2} - 4 \cdot 2 (- 1)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 1 + 3}{2 \cdot 2}, u_{2} = \frac{- 1 - 3}{2 \cdot 2}

u = \frac{- 1 + 3}{2 \cdot 2} : \frac{1}{2}

\frac{- 1 + 3}{2 \cdot 2}

Add/Subtract the numbers:

- 1 + 3 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 1 - 3}{2 \cdot 2} : - 1

\frac{- 1 - 3}{2 \cdot 2}

Subtract the numbers:

- 1 - 3 = - 4

= \frac{- 4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = - 1

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{2}, sin (θ) = - 1

sin (θ) = \frac{1}{2}, sin (θ) = - 1

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

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) = - 1 : θ = \frac{3 π}{2} + 2 πn

sin (θ) = - 1

General solutions for

sin (θ) = - 1

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{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Since the equation is undefined for:

\frac{3 π}{2} + 2 πn

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

Popular Examples

cos(x)=0.68

cos (x) = 0.68

cos(x)=0.65

cos (x) = 0.65

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

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

cos(2x)=(cos^2(x)+(sin^2(x)))

cos (2 x) = (cos^{2} (x) + (sin^{2} (x)))

tan(θ)=30.48(80)

tan (θ) = 30.48 (80)

Frequently Asked Questions (FAQ)

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