What is the general solution for cos(θ)+2tan(θ)=sec(θ) ?

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

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cos(θ)+2tan(θ)=sec(θ)

Solution

cos (θ) + 2 tan (θ) = sec (θ)

Solution

θ = 2 πn, θ = π + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (θ) + 2 tan (θ) = sec (θ)

Subtract

sec (θ)

from both sides

cos (θ) + 2 tan (θ) - sec (θ) = 0

Express with sin, cos

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

Simplify

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

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

Multiply

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

2 \cdot \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) \cdot 2}{cos ( θ )}

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

Combine the fractions

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

Apply rule

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

= \frac{2 sin ( θ ) - 1}{cos ( θ )}

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

cos (θ) cos (θ) + sin (θ) \cdot 2 - 1 = cos^{2} (θ) + 2 sin (θ) - 1

cos (θ) cos (θ) + sin (θ) \cdot 2 - 1

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) + 2 sin (θ) - 1

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

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

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

cos^{2} (θ) + 2 sin (θ) - 1 = 0

Subtract

2 sin (θ)

from both sides

cos^{2} (θ) - 1 = - 2 sin (θ)

Square both sides

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

Subtract

(- 2 sin (θ))^{2}

from both sides

(cos^{2} (θ) - 1)^{2} - 4 sin^{2} (θ) = 0

Factor

(cos^{2} (θ) - 1)^{2} - 4 sin^{2} (θ) : (cos^{2} (θ) - 1 + 2 sin (θ)) (cos^{2} (θ) - 1 - 2 sin (θ))

(cos^{2} (θ) - 1)^{2} - 4 sin^{2} (θ)

Rewrite

(cos^{2} (θ) - 1)^{2} - 4 sin^{2} (θ)

(cos^{2} (θ) - 1)^{2} - (2 sin (θ))^{2}

(cos^{2} (θ) - 1)^{2} - 4 sin^{2} (θ)

Rewrite

4

2^{2}

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

Apply exponent rule:

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

2^{2} sin^{2} (θ) = (2 sin (θ))^{2}

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

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

Apply Difference of Two Squares Formula:

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

(cos^{2} (θ) - 1)^{2} - (2 sin (θ))^{2} = ((cos^{2} (θ) - 1) + 2 sin (θ)) ((cos^{2} (θ) - 1) - 2 sin (θ))

= ((cos^{2} (θ) - 1) + 2 sin (θ)) ((cos^{2} (θ) - 1) - 2 sin (θ))

Refine

= (cos^{2} (θ) + 2 sin (θ) - 1) (cos^{2} (θ) - 2 sin (θ) - 1)

(cos^{2} (θ) - 1 + 2 sin (θ)) (cos^{2} (θ) - 1 - 2 sin (θ)) = 0

Solving each part separately

cos^{2} (θ) - 1 + 2 sin (θ) = 0 or cos^{2} (θ) - 1 - 2 sin (θ) = 0

cos^{2} (θ) - 1 + 2 sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

cos^{2} (θ) - 1 + 2 sin (θ) = 0

Rewrite using trig identities

- 1 + cos^{2} (θ) + 2 sin (θ)

Use the Pythagorean identity:

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

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

= 2 sin (θ) - sin^{2} (θ)

- sin^{2} (θ) + 2 sin (θ) = 0

Solve by substitution

- sin^{2} (θ) + 2 sin (θ) = 0

Let:

sin (θ) = u

- u^{2} + 2 u = 0

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

- u^{2} + 2 u = 0

Solve with the quadratic formula

- u^{2} + 2 u = 0

Quadratic Equation Formula:

For

a = - 1, b = 2, c = 0

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

2^{2} - 4 (- 1) \cdot 0 = 2

2^{2} - 4 (- 1) \cdot 0

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} + 0

2^{2} + 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 2 + 2 = 0

= \frac{0}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{- 2}

Apply the fraction rule:

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

= - \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 2 - 2 = - 4

= \frac{- 4}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{- 2}

Apply the fraction rule:

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

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

The solutions to the quadratic equation are:

u = 0, u = 2

Substitute back

u = sin (θ)

sin (θ) = 0, sin (θ) = 2

sin (θ) = 0, sin (θ) = 2

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 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 = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

sin (θ) = 2 :

No Solution

sin (θ) = 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

cos^{2} (θ) - 1 - 2 sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

cos^{2} (θ) - 1 - 2 sin (θ) = 0

Rewrite using trig identities

- 1 + cos^{2} (θ) - 2 sin (θ)

Use the Pythagorean identity:

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

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

= - 2 sin (θ) - sin^{2} (θ)

- sin^{2} (θ) - 2 sin (θ) = 0

Solve by substitution

- sin^{2} (θ) - 2 sin (θ) = 0

Let:

sin (θ) = u

- u^{2} - 2 u = 0

- u^{2} - 2 u = 0 : u = - 2, u = 0

- u^{2} - 2 u = 0

Solve with the quadratic formula

- u^{2} - 2 u = 0

Quadratic Equation Formula:

For

a = - 1, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

(- 2)^{2} - 4 (- 1) \cdot 0 = 2

(- 2)^{2} - 4 (- 1) \cdot 0

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 1 \cdot 0

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} + 0

2^{2} + 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Remove parentheses:

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

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

Add the numbers:

2 + 2 = 4

= \frac{4}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{- 2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

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

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

Remove parentheses:

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

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

Subtract the numbers:

2 - 2 = 0

= \frac{0}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{- 2}

Apply the fraction rule:

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

= - \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

The solutions to the quadratic equation are:

u = - 2, u = 0

Substitute back

u = sin (θ)

sin (θ) = - 2, sin (θ) = 0

sin (θ) = - 2, sin (θ) = 0

sin (θ) = - 2 :

No Solution

sin (θ) = - 2

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 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 = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

cos (θ) + 2 tan (θ) = sec (θ)

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

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

cos (θ) + 2 tan (θ) = sec (θ)

plug in

θ = 2 π 1

cos (2 π 1) + 2 tan (2 π 1) = sec (2 π 1)

Refine

1 = 1

\Rightarrow True

Check the solution

π + 2 πn :

True

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

cos (θ) + 2 tan (θ) = sec (θ)

plug in

θ = π + 2 π 1

cos (π + 2 π 1) + 2 tan (π + 2 π 1) = sec (π + 2 π 1)

Refine

- 1 = - 1

\Rightarrow True

θ = 2 πn, θ = π + 2 πn

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Frequently Asked Questions (FAQ)

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