What is the general solution for tan(2x)+sec(2x)=4 ?

The general solution for tan(2x)+sec(2x)=4 is x=(1.08083…)/2+pin

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tan(2x)+sec(2x)=4

Solution

tan (2 x) + sec (2 x) = 4

Solution

x = \frac{1.08083 \dots}{2} + πn

Degrees

x = 30.96375 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan (2 x) + sec (2 x) = 4

Subtract

4

from both sides

tan (2 x) + sec (2 x) - 4 = 0

Express with sin, cos

\frac{sin ( 2 x )}{cos ( 2 x )} + \frac{1}{cos ( 2 x )} - 4 = 0

Simplify

\frac{sin ( 2 x )}{cos ( 2 x )} + \frac{1}{cos ( 2 x )} - 4 : \frac{sin ( 2 x ) + 1 - 4 cos ( 2 x )}{cos ( 2 x )}

\frac{sin ( 2 x )}{cos ( 2 x )} + \frac{1}{cos ( 2 x )} - 4

Combine the fractions

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

Apply rule

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

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

= \frac{sin ( 2 x ) + 1}{cos ( 2 x )} - 4

Convert element to fraction:

4 = \frac{4 cos ( 2 x )}{cos ( 2 x )}

= \frac{sin ( 2 x ) + 1}{cos ( 2 x )} - \frac{4 cos ( 2 x )}{cos ( 2 x )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( 2 x ) + 1 - 4 cos ( 2 x )}{cos ( 2 x )}

\frac{sin ( 2 x ) + 1 - 4 cos ( 2 x )}{cos ( 2 x )} = 0

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

sin (2 x) + 1 - 4 cos (2 x) = 0

Add

4 cos (2 x)

to both sides

sin (2 x) + 1 = 4 cos (2 x)

Square both sides

(sin (2 x) + 1)^{2} = (4 cos (2 x))^{2}

Subtract

(4 cos (2 x))^{2}

from both sides

(sin (2 x) + 1)^{2} - 16 cos^{2} (2 x) = 0

Rewrite using trig identities

(1 + sin (2 x))^{2} - 16 cos^{2} (2 x)

Use the Pythagorean identity:

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

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

= (1 + sin (2 x))^{2} - 16 (1 - sin^{2} (2 x))

Simplify

(1 + sin (2 x))^{2} - 16 (1 - sin^{2} (2 x)) : 17 sin^{2} (2 x) + 2 sin (2 x) - 15

(1 + sin (2 x))^{2} - 16 (1 - sin^{2} (2 x))

(1 + sin (2 x))^{2} : 1 + 2 sin (2 x) + sin^{2} (2 x)

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 1, b = sin (2 x)

= 1^{2} + 2 \cdot 1 \cdot sin (2 x) + sin^{2} (2 x)

Simplify

1^{2} + 2 \cdot 1 \cdot sin (2 x) + sin^{2} (2 x) : 1 + 2 sin (2 x) + sin^{2} (2 x)

1^{2} + 2 \cdot 1 \cdot sin (2 x) + sin^{2} (2 x)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot sin (2 x) + sin^{2} (2 x)

Multiply the numbers:

2 \cdot 1 = 2

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

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

= 1 + 2 sin (2 x) + sin^{2} (2 x) - 16 (1 - sin^{2} (2 x))

Expand

- 16 (1 - sin^{2} (2 x)) : - 16 + 16 sin^{2} (2 x)

- 16 (1 - sin^{2} (2 x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 16, b = 1, c = sin^{2} (2 x)

= - 16 \cdot 1 - (- 16) sin^{2} (2 x)

Apply minus-plus rules

- (- a) = a

= - 16 \cdot 1 + 16 sin^{2} (2 x)

Multiply the numbers:

16 \cdot 1 = 16

= - 16 + 16 sin^{2} (2 x)

= 1 + 2 sin (2 x) + sin^{2} (2 x) - 16 + 16 sin^{2} (2 x)

Simplify

1 + 2 sin (2 x) + sin^{2} (2 x) - 16 + 16 sin^{2} (2 x) : 17 sin^{2} (2 x) + 2 sin (2 x) - 15

1 + 2 sin (2 x) + sin^{2} (2 x) - 16 + 16 sin^{2} (2 x)

Group like terms

= 2 sin (2 x) + sin^{2} (2 x) + 16 sin^{2} (2 x) + 1 - 16

Add similar elements:

sin^{2} (2 x) + 16 sin^{2} (2 x) = 17 sin^{2} (2 x)

= 2 sin (2 x) + 17 sin^{2} (2 x) + 1 - 16

Add/Subtract the numbers:

1 - 16 = - 15

= 17 sin^{2} (2 x) + 2 sin (2 x) - 15

= 17 sin^{2} (2 x) + 2 sin (2 x) - 15

= 17 sin^{2} (2 x) + 2 sin (2 x) - 15

- 15 + 17 sin^{2} (2 x) + 2 sin (2 x) = 0

Solve by substitution

- 15 + 17 sin^{2} (2 x) + 2 sin (2 x) = 0

Let:

sin (2 x) = u

- 15 + 17 u^{2} + 2 u = 0

- 15 + 17 u^{2} + 2 u = 0 : u = \frac{15}{17}, u = - 1

- 15 + 17 u^{2} + 2 u = 0

Write in the standard form

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

17 u^{2} + 2 u - 15 = 0

Solve with the quadratic formula

17 u^{2} + 2 u - 15 = 0

Quadratic Equation Formula:

For

a = 17, b = 2, c = - 15

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 17 ( - 15 )}{2 \cdot 17}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 17 ( - 15 )}{2 \cdot 17}

2^{2} - 4 \cdot 17 (- 15) = 32

2^{2} - 4 \cdot 17 (- 15)

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 17 \cdot 15

Multiply the numbers:

4 \cdot 17 \cdot 15 = 1020

= 2^{2} + 1020

2^{2} = 4

= 4 + 1020

Add the numbers:

4 + 1020 = 1024

= 1024

Factor the number:

1024 = 3 2^{2}

= 3 2^{2}

Apply radical rule:

3 2^{2} = 32

= 32

u_{1, 2} = \frac{- 2 \pm 32}{2 \cdot 17}

Separate the solutions

u_{1} = \frac{- 2 + 32}{2 \cdot 17}, u_{2} = \frac{- 2 - 32}{2 \cdot 17}

u = \frac{- 2 + 32}{2 \cdot 17} : \frac{15}{17}

\frac{- 2 + 32}{2 \cdot 17}

Add/Subtract the numbers:

- 2 + 32 = 30

= \frac{30}{2 \cdot 17}

Multiply the numbers:

2 \cdot 17 = 34

= \frac{30}{34}

Cancel the common factor:

2

= \frac{15}{17}

u = \frac{- 2 - 32}{2 \cdot 17} : - 1

\frac{- 2 - 32}{2 \cdot 17}

Subtract the numbers:

- 2 - 32 = - 34

= \frac{- 34}{2 \cdot 17}

Multiply the numbers:

2 \cdot 17 = 34

= \frac{- 34}{34}

Apply the fraction rule:

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

= - \frac{34}{34}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{15}{17}, u = - 1

Substitute back

u = sin (2 x)

sin (2 x) = \frac{15}{17}, sin (2 x) = - 1

sin (2 x) = \frac{15}{17}, sin (2 x) = - 1

sin (2 x) = \frac{15}{17} : x = \frac{arcsin ( \frac{15}{17} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn

sin (2 x) = \frac{15}{17}

Apply trig inverse properties

sin (2 x) = \frac{15}{17}

General solutions for

sin (2 x) = \frac{15}{17}

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

2 x = arcsin (\frac{15}{17}) + 2 πn, 2 x = π - arcsin (\frac{15}{17}) + 2 πn

2 x = arcsin (\frac{15}{17}) + 2 πn, 2 x = π - arcsin (\frac{15}{17}) + 2 πn

Solve

2 x = arcsin (\frac{15}{17}) + 2 πn : x = \frac{arcsin ( \frac{15}{17} )}{2} + πn

2 x = arcsin (\frac{15}{17}) + 2 πn

Divide both sides by

2

2 x = arcsin (\frac{15}{17}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arcsin ( \frac{15}{17} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{arcsin ( \frac{15}{17} )}{2} + πn

x = \frac{arcsin ( \frac{15}{17} )}{2} + πn

Solve

2 x = π - arcsin (\frac{15}{17}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn

2 x = π - arcsin (\frac{15}{17}) + 2 πn

Divide both sides by

2

2 x = π - arcsin (\frac{15}{17}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn

x = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn

x = \frac{arcsin ( \frac{15}{17} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn

sin (2 x) = - 1 : x = \frac{3 π}{4} + πn

sin (2 x) = - 1

General solutions for

sin (2 x) = - 1

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}

2 x = \frac{3 π}{2} + 2 πn

2 x = \frac{3 π}{2} + 2 πn

Solve

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

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

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

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{arcsin ( \frac{15}{17} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn, x = \frac{3 π}{4} + πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

tan (2 x) + sec (2 x) = 4

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

Check the solution

\frac{arcsin ( \frac{15}{17} )}{2} + πn :

True

\frac{arcsin ( \frac{15}{17} )}{2} + πn

Plug in

n = 1

\frac{arcsin ( \frac{15}{17} )}{2} + π 1

For

tan (2 x) + sec (2 x) = 4

plug in

x = \frac{arcsin ( \frac{15}{17} )}{2} + π 1

tan (2 (\frac{arcsin ( \frac{15}{17} )}{2} + π 1)) + sec (2 (\frac{arcsin ( \frac{15}{17} )}{2} + π 1)) = 4

Refine

4 = 4

\Rightarrow True

Check the solution

\frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn :

False

\frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn

Plug in

n = 1

\frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + π 1

For

tan (2 x) + sec (2 x) = 4

plug in

x = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + π 1

tan (2 (\frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + π 1)) + sec (2 (\frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + π 1)) = 4

Refine

- 4 = 4

\Rightarrow False

Check the solution

\frac{3 π}{4} + πn :

False

\frac{3 π}{4} + πn

Plug in

n = 1

\frac{3 π}{4} + π 1

For

tan (2 x) + sec (2 x) = 4

plug in

x = \frac{3 π}{4} + π 1

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

Undefined

\Rightarrow False

x = \frac{arcsin ( \frac{15}{17} )}{2} + πn

Show solutions in decimal form

x = \frac{1.08083 \dots}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan(2x)+sec(2x)=4 ?
The general solution for tan(2x)+sec(2x)=4 is x=(1.08083…)/2+pin