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Popular >

tan(θ^2)+tan(θ)=0

Solution

tan (θ^{2}) + tan (θ) = 0

Solution

θ = \frac{- 1 + 1 + 8 πn}{2}, θ = \frac{- 1 - 1 + 8 πn}{2}, θ = \frac{- 1 + 1 - 4 ( - 2 πn - π )}{2}, θ = \frac{- 1 - 1 - 4 ( - 2 πn - π )}{2}

Degrees

θ = - 28.64788 \dots^{\circ} + 146.44856 \dots^{\circ} n, θ = - 28.64788 \dots^{\circ} - 146.44856 \dots^{\circ} n, θ = - 28.64788 \dots^{\circ} + 178.21454 \dots^{\circ} n, θ = - 28.64788 \dots^{\circ} - 178.21454 \dots^{\circ} n

Solution steps

tan (θ^{2}) + tan (θ) = 0

Express with sin, cos

tan (θ) + tan (θ^{2})

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Least Common Multiplier of

cos (θ), cos (θ^{2}) : cos (θ) cos (θ^{2})

cos (θ), cos (θ^{2})

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (θ)

cos (θ^{2})

= cos (θ) cos (θ^{2})

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

cos (θ) cos (θ^{2})

For

\frac{sin ( θ )}{cos ( θ )} :

multiply the denominator and numerator by

cos (θ^{2})

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

For

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

multiply the denominator and numerator by

cos (θ)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

Rewrite using trig identities

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

Use the Angle Sum identity:

sin (s) cos (t) + cos (s) sin (t) = sin (s + t)

= sin (θ + θ^{2})

sin (θ + θ^{2}) = 0

General solutions for

sin (θ + θ^{2}) = 0

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 + θ^{2} = 0 + 2 πn, θ + θ^{2} = π + 2 πn

θ + θ^{2} = 0 + 2 πn, θ + θ^{2} = π + 2 πn

Solve

θ + θ^{2} = 0 + 2 πn : θ = \frac{- 1 + 1 + 8 πn}{2}, θ = \frac{- 1 - 1 + 8 πn}{2}

θ + θ^{2} = 0 + 2 πn

Expand

0 + 2 πn : 2 πn

0 + 2 πn

0 + 2 πn = 2 πn

= 2 πn

θ + θ^{2} = 2 πn

Move

2 πn

to the left side

θ + θ^{2} = 2 πn

Subtract

2 πn

from both sides

θ + θ^{2} - 2 πn = 2 πn - 2 πn

Simplify

θ + θ^{2} - 2 πn = 0

θ + θ^{2} - 2 πn = 0

Write in the standard form

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

θ^{2} + θ - 2 πn = 0

Solve with the quadratic formula

θ^{2} + θ - 2 πn = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = - 2 πn

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

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

Simplify

1^{2} - 4 \cdot 1 \cdot (- 2 πn) : 1 + 8 πn

1^{2} - 4 \cdot 1 \cdot (- 2 πn)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 2 πn)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2 πn

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 1 + 8 πn

θ_{1, 2} = \frac{- 1 \pm 1 + 8 πn}{2 \cdot 1}

Separate the solutions

θ_{1} = \frac{- 1 + 1 + 8 πn}{2 \cdot 1}, θ_{2} = \frac{- 1 - 1 + 8 πn}{2 \cdot 1}

θ = \frac{- 1 + 1 + 8 πn}{2 \cdot 1} : \frac{- 1 + 1 + 8 πn}{2}

\frac{- 1 + 1 + 8 πn}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 8 πn + 1}{2}

θ = \frac{- 1 - 1 + 8 πn}{2 \cdot 1} : \frac{- 1 - 1 + 8 πn}{2}

\frac{- 1 - 1 + 8 πn}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 8 πn + 1}{2}

The solutions to the quadratic equation are:

θ = \frac{- 1 + 1 + 8 πn}{2}, θ = \frac{- 1 - 1 + 8 πn}{2}

Solve

θ + θ^{2} = π + 2 πn : θ = \frac{- 1 + 1 - 4 ( - 2 πn - π )}{2}, θ = \frac{- 1 - 1 - 4 ( - 2 πn - π )}{2}

θ + θ^{2} = π + 2 πn

Move

2 πn

to the left side

θ + θ^{2} = π + 2 πn

Subtract

2 πn

from both sides

θ + θ^{2} - 2 πn = π + 2 πn - 2 πn

Simplify

θ + θ^{2} - 2 πn = π

θ + θ^{2} - 2 πn = π

Move

π

to the left side

θ + θ^{2} - 2 πn = π

Subtract

π

from both sides

θ + θ^{2} - 2 πn - π = π - π

Simplify

θ + θ^{2} - 2 πn - π = 0

θ + θ^{2} - 2 πn - π = 0

Write in the standard form

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

θ^{2} + θ - 2 πn - π = 0

Solve with the quadratic formula

θ^{2} + θ - 2 πn - π = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = - 2 πn - π

θ_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 2 πn - π )}{2 \cdot 1}

θ_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 2 πn - π )}{2 \cdot 1}

Simplify

1^{2} - 4 \cdot 1 \cdot (- 2 πn - π) : 1 - 4 (- 2 πn - π)

1^{2} - 4 \cdot 1 \cdot (- 2 πn - π)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 2 πn - π)

Multiply the numbers:

4 \cdot 1 = 4

= 1 - 4 (- 2 πn - π)

θ_{1, 2} = \frac{- 1 \pm 1 - 4 ( - 2 πn - π )}{2 \cdot 1}

Separate the solutions

θ_{1} = \frac{- 1 + 1 - 4 ( - 2 πn - π )}{2 \cdot 1}, θ_{2} = \frac{- 1 - 1 - 4 ( - 2 πn - π )}{2 \cdot 1}

θ = \frac{- 1 + 1 - 4 ( - 2 πn - π )}{2 \cdot 1} : \frac{- 1 + 1 - 4 ( - 2 πn - π )}{2}

\frac{- 1 + 1 - 4 ( - 2 πn - π )}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + - 4 ( - 2 πn - π ) + 1}{2}

θ = \frac{- 1 - 1 - 4 ( - 2 πn - π )}{2 \cdot 1} : \frac{- 1 - 1 - 4 ( - 2 πn - π )}{2}

\frac{- 1 - 1 - 4 ( - 2 πn - π )}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - - 4 ( - 2 πn - π ) + 1}{2}

The solutions to the quadratic equation are:

θ = \frac{- 1 + 1 - 4 ( - 2 πn - π )}{2}, θ = \frac{- 1 - 1 - 4 ( - 2 πn - π )}{2}

θ = \frac{- 1 + 1 + 8 πn}{2}, θ = \frac{- 1 - 1 + 8 πn}{2}, θ = \frac{- 1 + 1 - 4 ( - 2 πn - π )}{2}, θ = \frac{- 1 - 1 - 4 ( - 2 πn - π )}{2}

Graph

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