What is the general solution for tan(θ)=(sqrt(2))/2 csc(θ) ?

The general solution for tan(θ)=(sqrt(2))/2 csc(θ) is θ= pi/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 Trigonometry >

tan(θ)=(sqrt(2))/2 csc(θ)

Solution

tan (θ) = \frac{2}{2} csc (θ)

Solution

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

Degrees

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

tan (θ) = \frac{2}{2} csc (θ)

Subtract

\frac{2}{2} csc (θ)

from both sides

tan (θ) - \frac{csc ( θ )}{2} = 0

Simplify

tan (θ) - \frac{csc ( θ )}{2} : \frac{2 tan ( θ ) - csc ( θ )}{2}

tan (θ) - \frac{csc ( θ )}{2}

Convert element to fraction:

tan (θ) = \frac{tan ( θ ) 2}{2}

= \frac{tan ( θ ) 2}{2} - \frac{csc ( θ )}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{tan ( θ ) 2 - csc ( θ )}{2}

\frac{2 tan ( θ ) - csc ( θ )}{2} = 0

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

2 tan (θ) - csc (θ) = 0

Express with sin, cos

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

Simplify

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

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

Least Common Multiplier of

cos (θ), sin (θ) : cos (θ) sin (θ)

cos (θ), sin (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (θ)

sin (θ)

= cos (θ) sin (θ)

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) sin (θ)

For

\frac{sin ( θ ) 2}{cos ( θ )} :

multiply the denominator and numerator by

sin (θ)

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

For

\frac{1}{sin ( θ )} :

multiply the denominator and numerator by

cos (θ)

\frac{1}{sin ( θ )} = \frac{1 \cdot cos ( θ )}{sin ( θ ) cos ( θ )} = \frac{cos ( θ )}{cos ( θ ) sin ( θ )}

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

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

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

2 sin^{2} (θ) - cos (θ) = 0

Add

cos (θ)

to both sides

2 sin^{2} (θ) = cos (θ)

Square both sides

(2 sin^{2} (θ))^{2} = cos^{2} (θ)

Subtract

cos^{2} (θ)

from both sides

2 sin^{4} (θ) - cos^{2} (θ) = 0

Factor

2 sin^{4} (θ) - cos^{2} (θ) : (2 sin^{2} (θ) + cos (θ)) (2 sin^{2} (θ) - cos (θ))

2 sin^{4} (θ) - cos^{2} (θ)

Rewrite

2 sin^{4} (θ) - cos^{2} (θ)

(2 sin^{2} (θ))^{2} - cos^{2} (θ)

2 sin^{4} (θ) - cos^{2} (θ)

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} sin^{4} (θ) - cos^{2} (θ)

Apply exponent rule:

a^{b c} = (a^{b})^{c}

sin^{4} (θ) = (sin^{2} (θ))^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

(2)^{2} (sin^{2} (θ))^{2} = (2 sin^{2} (θ))^{2}

= (2 sin^{2} (θ))^{2} - cos^{2} (θ)

= (2 sin^{2} (θ))^{2} - cos^{2} (θ)

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

Solving each part separately

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

2 sin^{2} (θ) + cos (θ) = 0 : θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

2 sin^{2} (θ) + cos (θ) = 0

Rewrite using trig identities

cos (θ) + sin^{2} (θ) 2

Use the Pythagorean identity:

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

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

= cos (θ) + (1 - cos^{2} (θ)) 2

cos (θ) + (1 - cos^{2} (θ)) 2 = 0

Solve by substitution

cos (θ) + (1 - cos^{2} (θ)) 2 = 0

Let:

cos (θ) = u

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

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

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

Expand

u + (1 - u^{2}) 2 : u + 2 - 2 u^{2}

u + (1 - u^{2}) 2

= u + 2 (1 - u^{2})

Expand

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 - 2 u^{2}

= 1 \cdot 2 - 2 u^{2}

Multiply:

1 \cdot 2 = 2

= 2 - 2 u^{2}

= u + 2 - 2 u^{2}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 42 (- 2)

Apply rule

- (- a) = a

= 1 + 422

422 = 8

422

Apply radical rule:

a a = a

22 = 2

= 4 \cdot 2

Multiply the numbers:

4 \cdot 2 = 8

= 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 ( - 2 )}

Separate the solutions

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

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

\frac{- 1 + 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 1 + 3 = 2

= \frac{2}{- 2 2}

Apply the fraction rule:

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

= - \frac{2}{2 2}

Divide the numbers:

\frac{2}{2} = 1

= - \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}

u = \frac{- 1 - 3}{2 ( - 2 )} : 2

\frac{- 1 - 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 1 - 3 = - 4

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

Apply the fraction rule:

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

= \frac{4}{2 2}

Divide the numbers:

\frac{4}{2} = 2

= \frac{2}{2}

Apply radical rule:

2 = 2^{\frac{1}{2}}

= \frac{2}{2 ^{\frac{1}{2}}}

Apply exponent rule:

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

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

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

Subtract the numbers:

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

= 2^{\frac{1}{2}}

Apply radical rule:

2^{\frac{1}{2}} = 2

= 2

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = - \frac{2}{2}, cos (θ) = 2

cos (θ) = - \frac{2}{2}, cos (θ) = 2

cos (θ) = - \frac{2}{2} : θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

cos (θ) = - \frac{2}{2}

General solutions for

cos (θ) = - \frac{2}{2}

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}

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

cos (θ) = 2 :

No Solution

cos (θ) = 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

2 sin^{2} (θ) - cos (θ) = 0 : θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

2 sin^{2} (θ) - cos (θ) = 0

Rewrite using trig identities

- cos (θ) + sin^{2} (θ) 2

Use the Pythagorean identity:

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

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

= - cos (θ) + (1 - cos^{2} (θ)) 2

- cos (θ) + (1 - cos^{2} (θ)) 2 = 0

Solve by substitution

- cos (θ) + (1 - cos^{2} (θ)) 2 = 0

Let:

cos (θ) = u

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

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

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

Expand

- u + (1 - u^{2}) 2 : - u + 2 - 2 u^{2}

- u + (1 - u^{2}) 2

= - u + 2 (1 - u^{2})

Expand

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 - 2 u^{2}

= 1 \cdot 2 - 2 u^{2}

Multiply:

1 \cdot 2 = 2

= 2 - 2 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

= (- 1)^{2} + 422

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

422 = 8

422

Apply radical rule:

a a = a

22 = 2

= 4 \cdot 2

Multiply the numbers:

4 \cdot 2 = 8

= 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 ( - 2 )}

Separate the solutions

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

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

\frac{- ( - 1 ) + 3}{2 ( - 2 )}

Remove parentheses:

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

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

Add the numbers:

1 + 3 = 4

= \frac{4}{- 2 2}

Apply the fraction rule:

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

= - \frac{4}{2 2}

Divide the numbers:

\frac{4}{2} = 2

= \frac{2}{2}

Apply radical rule:

2 = 2^{\frac{1}{2}}

= \frac{2}{2 ^{\frac{1}{2}}}

Apply exponent rule:

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

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

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

Subtract the numbers:

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

= 2^{\frac{1}{2}}

Apply radical rule:

2^{\frac{1}{2}} = 2

= - 2

u = \frac{- ( - 1 ) - 3}{2 ( - 2 )} : \frac{2}{2}

\frac{- ( - 1 ) - 3}{2 ( - 2 )}

Remove parentheses:

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

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

Subtract the numbers:

1 - 3 = - 2

= \frac{- 2}{- 2 2}

Apply the fraction rule:

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

= \frac{2}{2 2}

Divide the numbers:

\frac{2}{2} = 1

= \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}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = - 2, cos (θ) = \frac{2}{2}

cos (θ) = - 2, cos (θ) = \frac{2}{2}

cos (θ) = - 2 :

No Solution

cos (θ) = - 2

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = \frac{2}{2} : θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

cos (θ) = \frac{2}{2}

General solutions for

cos (θ) = \frac{2}{2}

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}

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

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

Combine all the solutions

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

Combine all the solutions

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn, θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

tan (θ) = \frac{2}{2} csc (θ)

Remove the ones that don't agree with the equation.

Check the solution

\frac{3 π}{4} + 2 πn :

False

\frac{3 π}{4} + 2 πn

Plug in

n = 1

\frac{3 π}{4} + 2 π 1

For

tan (θ) = \frac{2}{2} csc (θ)

plug in

θ = \frac{3 π}{4} + 2 π 1

tan (\frac{3 π}{4} + 2 π 1) = \frac{2}{2} csc (\frac{3 π}{4} + 2 π 1)

Refine

- 1 = 1

\Rightarrow False

Check the solution

\frac{5 π}{4} + 2 πn :

False

\frac{5 π}{4} + 2 πn

Plug in

n = 1

\frac{5 π}{4} + 2 π 1

For

tan (θ) = \frac{2}{2} csc (θ)

plug in

θ = \frac{5 π}{4} + 2 π 1

tan (\frac{5 π}{4} + 2 π 1) = \frac{2}{2} csc (\frac{5 π}{4} + 2 π 1)

Refine

1 = - 1

\Rightarrow False

Check the solution

\frac{π}{4} + 2 πn :

True

\frac{π}{4} + 2 πn

Plug in

n = 1

\frac{π}{4} + 2 π 1

For

tan (θ) = \frac{2}{2} csc (θ)

plug in

θ = \frac{π}{4} + 2 π 1

tan (\frac{π}{4} + 2 π 1) = \frac{2}{2} csc (\frac{π}{4} + 2 π 1)

Refine

1 = 1

\Rightarrow True

Check the solution

\frac{7 π}{4} + 2 πn :

True

\frac{7 π}{4} + 2 πn

Plug in

n = 1

\frac{7 π}{4} + 2 π 1

For

tan (θ) = \frac{2}{2} csc (θ)

plug in

θ = \frac{7 π}{4} + 2 π 1

tan (\frac{7 π}{4} + 2 π 1) = \frac{2}{2} csc (\frac{7 π}{4} + 2 π 1)

Refine

- 1 = - 1

\Rightarrow True

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

Graph

Popular Examples

Frequently Asked Questions (FAQ)

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