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

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

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Popular >

cos(θ)=tan(θ)

Solution

cos (θ) = tan (θ)

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

cos (θ) = tan (θ)

Subtract

tan (θ)

from both sides

cos (θ) - tan (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 (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{5 - 1}{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 ( - 1 ) \cdot 1}{2 ( - 1 )}

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

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

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

Apply rule

- (- a) = a

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

(- 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

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

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

Separate the solutions

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

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

\frac{- ( - 1 ) + 5}{2 ( - 1 )}

Remove parentheses:

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

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

u = \frac{- ( - 1 ) - 5}{2 ( - 1 )} : \frac{5 - 1}{2}

\frac{- ( - 1 ) - 5}{2 ( - 1 )}

Remove parentheses:

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

1 - 5 = - (5 - 1)

= \frac{5 - 1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

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

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

sin (θ) = - \frac{1 + 5}{2} :

No Solution

sin (θ) = - \frac{1 + 5}{2}

- 1 \leq sin (x) \leq 1

No Solution

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

sin (θ) = \frac{5 - 1}{2}

Apply trig inverse properties

sin (θ) = \frac{5 - 1}{2}

General solutions for

sin (θ) = \frac{5 - 1}{2}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

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

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Frequently Asked Questions (FAQ)

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