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

The general solution for tan(θ)=(cos(θ))/(sin(θ)) is θ= pi/4+pin,θ=(3pi)/4+pin

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tan(θ)=(cos(θ))/(sin(θ))

Solution

tan (θ) = \frac{cos ( θ )}{sin ( θ )}

Solution

θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

tan (θ) = \frac{cos ( θ )}{sin ( θ )}

Subtract

\frac{cos ( θ )}{sin ( θ )}

from both sides

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

Simplify

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

tan (θ) - \frac{cos ( θ )}{sin ( θ )}

Convert element to fraction:

tan (θ) = \frac{tan ( θ ) sin ( θ )}{sin ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

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

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

tan (θ) sin (θ) - cos (θ) = 0

Express with sin, cos

- cos (θ) + sin (θ) tan (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

sin (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

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

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= \frac{sin ^{2} ( θ )}{cos ( θ )}

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

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

- cos (θ) cos (θ) + sin^{2} (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^{2} (θ)

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

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

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

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

- cos^{2} (θ) + sin^{2} (θ) = 0

Rewrite using trig identities

- cos^{2} (θ) + sin^{2} (θ)

Use the Double Angle identity:

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

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

= - cos (2 θ)

- cos (2 θ) = 0

Divide both sides by

- 1

- cos (2 θ) = 0

Divide both sides by

- 1

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

Simplify

cos (2 θ) = 0

cos (2 θ) = 0

General solutions for

cos (2 θ) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Solve

2 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{4} + πn

2 θ = \frac{π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{π}{2} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Solve

2 θ = \frac{3 π}{2} + 2 πn : θ = \frac{3 π}{4} + πn

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

Popular Examples

sin(θ)+2sin^2(θ)=0

sin (θ) + 2 sin^{2} (θ) = 0

(cot(x)+1)(sin(x)+1)=0

(cot (x) + 1) (sin (x) + 1) = 0

cos^2(x)-2sin(x)=0

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

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

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

sin(3x)-3sin(x)=0

sin (3 x) - 3 sin (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for tan(θ)=(cos(θ))/(sin(θ)) ?
The general solution for tan(θ)=(cos(θ))/(sin(θ)) is θ= pi/4+pin,θ=(3pi)/4+pin