What is the general solution for (tan(θ))/(cos(θ))=1 ?

The general solution for (tan(θ))/(cos(θ))=1 is θ=0.66623…+2pin,θ=pi-0.66623…+2pin

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Popular >

(tan(θ))/(cos(θ))=1

Solution

\frac{tan ( θ )}{cos ( θ )} = 1

Solution

θ = 0.66623 \dots + 2 πn, θ = π - 0.66623 \dots + 2 πn

Degrees

θ = 38.17270 \dots^{\circ} + 36 0^{\circ} n, θ = 141.82729 \dots^{\circ} + 36 0^{\circ} n

Solution steps

\frac{tan ( θ )}{cos ( θ )} = 1

Subtract

1

from both sides

\frac{tan ( θ )}{cos ( θ )} - 1 = 0

Simplify

\frac{tan ( θ )}{cos ( θ )} - 1 : \frac{tan ( θ ) - cos ( θ )}{cos ( θ )}

\frac{tan ( θ )}{cos ( θ )} - 1

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{tan ( θ ) - cos ( θ )}{cos ( θ )}

\frac{tan ( θ ) - cos ( θ )}{cos ( θ )} = 0

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

tan (θ) - cos (θ) = 0

Express with sin, cos

- cos (θ) + tan (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (θ) cos (θ) + sin (θ)

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= - cos^{2} (θ) + sin (θ)

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

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

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

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

- cos^{2} (θ) + sin (θ) = 0

Rewrite using trig identities

- cos^{2} (θ) + sin (θ)

Use the Pythagorean identity:

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

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

= - (1 - sin^{2} (θ)) + sin (θ)

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

- (1 - sin^{2} (θ))

Distribute parentheses

= - (1) - (- sin^{2} (θ))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + sin^{2} (θ)

= - 1 + sin^{2} (θ) + sin (θ)

- 1 + sin (θ) + sin^{2} (θ) = 0

Solve by substitution

- 1 + sin (θ) + sin^{2} (θ) = 0

Let:

sin (θ) = u

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

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

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

Write in the standard form

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

u^{2} + u - 1 = 0

Solve with the quadratic formula

u^{2} + u - 1 = 0

Quadratic Equation Formula:

For

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

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

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

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 5}{2}

The solutions to the quadratic equation are:

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

Substitute back

u = sin (θ)

sin (θ) = \frac{- 1 + 5}{2}, sin (θ) = \frac{- 1 - 5}{2}

sin (θ) = \frac{- 1 + 5}{2}, sin (θ) = \frac{- 1 - 5}{2}

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

sin (θ) = \frac{- 1 + 5}{2}

Apply trig inverse properties

sin (θ) = \frac{- 1 + 5}{2}

General solutions for

sin (θ) = \frac{- 1 + 5}{2}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (\frac{- 1 + 5}{2}) + 2 πn, θ = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

θ = arcsin (\frac{- 1 + 5}{2}) + 2 πn, θ = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

sin (θ) = \frac{- 1 - 5}{2} :

No Solution

sin (θ) = \frac{- 1 - 5}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

θ = arcsin (\frac{- 1 + 5}{2}) + 2 πn, θ = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

Show solutions in decimal form

θ = 0.66623 \dots + 2 πn, θ = π - 0.66623 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for (tan(θ))/(cos(θ))=1 ?
The general solution for (tan(θ))/(cos(θ))=1 is θ=0.66623…+2pin,θ=pi-0.66623…+2pin