What is the general solution for sec(2x)+tan(2x)= 1/2 ?

The general solution for sec(2x)+tan(2x)= 1/2 is x=-(0.64350…)/2+pin

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Popular Trigonometry >

sec(2x)+tan(2x)= 1/2

Solution

sec (2 x) + tan (2 x) = \frac{1}{2}

Solution

x = - \frac{0.64350 \dots}{2} + πn

Degrees

x = - 18.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

sec (2 x) + tan (2 x) = \frac{1}{2}

Subtract

\frac{1}{2}

from both sides

sec (2 x) + tan (2 x) - \frac{1}{2} = 0

Simplify

sec (2 x) + tan (2 x) - \frac{1}{2} : \frac{2 sec ( 2 x ) + 2 tan ( 2 x ) - 1}{2}

sec (2 x) + tan (2 x) - \frac{1}{2}

Convert element to fraction:

sec (2 x) = \frac{sec ( 2 x ) 2}{2}, tan (2 x) = \frac{tan ( 2 x ) 2}{2}

= \frac{sec ( 2 x ) \cdot 2}{2} + \frac{tan ( 2 x ) \cdot 2}{2} - \frac{1}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{sec ( 2 x ) \cdot 2 + tan ( 2 x ) \cdot 2 - 1}{2}

\frac{2 sec ( 2 x ) + 2 tan ( 2 x ) - 1}{2} = 0

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

2 sec (2 x) + 2 tan (2 x) - 1 = 0

Express with sin, cos

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

Simplify

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

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

2 \cdot \frac{1}{cos ( 2 x )} = \frac{2}{cos ( 2 x )}

2 \cdot \frac{1}{cos ( 2 x )}

Multiply fractions:

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

= \frac{1 \cdot 2}{cos ( 2 x )}

Multiply the numbers:

1 \cdot 2 = 2

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

2 \cdot \frac{sin ( 2 x )}{cos ( 2 x )} = \frac{2 sin ( 2 x )}{cos ( 2 x )}

2 \cdot \frac{sin ( 2 x )}{cos ( 2 x )}

Multiply fractions:

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

= \frac{sin ( 2 x ) \cdot 2}{cos ( 2 x )}

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

Combine the fractions

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

Apply rule

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

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

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

Convert element to fraction:

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

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

Multiply:

1 \cdot cos (2 x) = cos (2 x)

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

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

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

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

Add

cos (2 x)

to both sides

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

Square both sides

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

Subtract

cos^{2} (2 x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

(2 + 2 sin (2 x))^{2} - (1 - sin^{2} (2 x)) : 5 sin^{2} (2 x) + 8 sin (2 x) + 3

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

(2 + 2 sin (2 x))^{2} : 4 + 8 sin (2 x) + 4 sin^{2} (2 x)

Apply Perfect Square Formula:

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

a = 2, b = 2 sin (2 x)

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

Simplify

2^{2} + 2 \cdot 2 \cdot 2 sin (2 x) + (2 sin (2 x))^{2} : 4 + 8 sin (2 x) + 4 sin^{2} (2 x)

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

2^{2} = 4

2^{2}

2^{2} = 4

= 4

2 \cdot 2 \cdot 2 sin (2 x) = 8 sin (2 x)

2 \cdot 2 \cdot 2 sin (2 x)

Multiply the numbers:

2 \cdot 2 \cdot 2 = 8

= 8 sin (2 x)

(2 sin (2 x))^{2} = 4 sin^{2} (2 x)

(2 sin (2 x))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

2^{2} = 4

= 4 sin^{2} (2 x)

= 4 + 8 sin (2 x) + 4 sin^{2} (2 x)

= 4 + 8 sin (2 x) + 4 sin^{2} (2 x)

= 4 + 8 sin (2 x) + 4 sin^{2} (2 x) - (1 - sin^{2} (2 x))

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

= 4 + 8 sin (2 x) + 4 sin^{2} (2 x) - 1 + sin^{2} (2 x)

Simplify

4 + 8 sin (2 x) + 4 sin^{2} (2 x) - 1 + sin^{2} (2 x) : 5 sin^{2} (2 x) + 8 sin (2 x) + 3

4 + 8 sin (2 x) + 4 sin^{2} (2 x) - 1 + sin^{2} (2 x)

Group like terms

= 8 sin (2 x) + 4 sin^{2} (2 x) + sin^{2} (2 x) + 4 - 1

Add similar elements:

4 sin^{2} (2 x) + sin^{2} (2 x) = 5 sin^{2} (2 x)

= 8 sin (2 x) + 5 sin^{2} (2 x) + 4 - 1

Add/Subtract the numbers:

4 - 1 = 3

= 5 sin^{2} (2 x) + 8 sin (2 x) + 3

= 5 sin^{2} (2 x) + 8 sin (2 x) + 3

= 5 sin^{2} (2 x) + 8 sin (2 x) + 3

3 + 5 sin^{2} (2 x) + 8 sin (2 x) = 0

Solve by substitution

3 + 5 sin^{2} (2 x) + 8 sin (2 x) = 0

Let:

sin (2 x) = u

3 + 5 u^{2} + 8 u = 0

3 + 5 u^{2} + 8 u = 0 : u = - \frac{3}{5}, u = - 1

3 + 5 u^{2} + 8 u = 0

Write in the standard form

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

5 u^{2} + 8 u + 3 = 0

Solve with the quadratic formula

5 u^{2} + 8 u + 3 = 0

Quadratic Equation Formula:

For

a = 5, b = 8, c = 3

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 5 \cdot 3}{2 \cdot 5}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 5 \cdot 3}{2 \cdot 5}

8^{2} - 4 \cdot 5 \cdot 3 = 2

8^{2} - 4 \cdot 5 \cdot 3

Multiply the numbers:

4 \cdot 5 \cdot 3 = 60

= 8^{2} - 60

8^{2} = 64

= 64 - 60

Subtract the numbers:

64 - 60 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u_{1, 2} = \frac{- 8 \pm 2}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- 8 + 2}{2 \cdot 5}, u_{2} = \frac{- 8 - 2}{2 \cdot 5}

u = \frac{- 8 + 2}{2 \cdot 5} : - \frac{3}{5}

\frac{- 8 + 2}{2 \cdot 5}

Add/Subtract the numbers:

- 8 + 2 = - 6

= \frac{- 6}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 6}{10}

Apply the fraction rule:

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

= - \frac{6}{10}

Cancel the common factor:

2

= - \frac{3}{5}

u = \frac{- 8 - 2}{2 \cdot 5} : - 1

\frac{- 8 - 2}{2 \cdot 5}

Subtract the numbers:

- 8 - 2 = - 10

= \frac{- 10}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 10}{10}

Apply the fraction rule:

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

= - \frac{10}{10}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = - \frac{3}{5}, u = - 1

Substitute back

u = sin (2 x)

sin (2 x) = - \frac{3}{5}, sin (2 x) = - 1

sin (2 x) = - \frac{3}{5}, sin (2 x) = - 1

sin (2 x) = - \frac{3}{5} : x = - \frac{arcsin ( \frac{3}{5} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + πn

sin (2 x) = - \frac{3}{5}

Apply trig inverse properties

sin (2 x) = - \frac{3}{5}

General solutions for

sin (2 x) = - \frac{3}{5}

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

2 x = arcsin (- \frac{3}{5}) + 2 πn, 2 x = π + arcsin (\frac{3}{5}) + 2 πn

2 x = arcsin (- \frac{3}{5}) + 2 πn, 2 x = π + arcsin (\frac{3}{5}) + 2 πn

Solve

2 x = arcsin (- \frac{3}{5}) + 2 πn : x = - \frac{arcsin ( \frac{3}{5} )}{2} + πn

2 x = arcsin (- \frac{3}{5}) + 2 πn

Simplify

arcsin (- \frac{3}{5}) + 2 πn : - arcsin (\frac{3}{5}) + 2 πn

arcsin (- \frac{3}{5}) + 2 πn

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{3}{5}) = - arcsin (\frac{3}{5})

= - arcsin (\frac{3}{5}) + 2 πn

2 x = - arcsin (\frac{3}{5}) + 2 πn

Divide both sides by

2

2 x = - arcsin (\frac{3}{5}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = - \frac{arcsin ( \frac{3}{5} )}{2} + \frac{2 πn}{2}

Simplify

x = - \frac{arcsin ( \frac{3}{5} )}{2} + πn

x = - \frac{arcsin ( \frac{3}{5} )}{2} + πn

Solve

2 x = π + arcsin (\frac{3}{5}) + 2 πn : x = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + πn

2 x = π + arcsin (\frac{3}{5}) + 2 πn

Divide both sides by

2

2 x = π + arcsin (\frac{3}{5}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + πn

x = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + πn

x = - \frac{arcsin ( \frac{3}{5} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{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{3}{5} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + πn, x = \frac{3 π}{4} + πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

sec (2 x) + tan (2 x) = \frac{1}{2}

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

Check the solution

- \frac{arcsin ( \frac{3}{5} )}{2} + πn :

True

- \frac{arcsin ( \frac{3}{5} )}{2} + πn

Plug in

n = 1

- \frac{arcsin ( \frac{3}{5} )}{2} + π 1

For

sec (2 x) + tan (2 x) = \frac{1}{2}

plug in

x = - \frac{arcsin ( \frac{3}{5} )}{2} + π 1

sec (2 (- \frac{arcsin ( \frac{3}{5} )}{2} + π 1)) + tan (2 (- \frac{arcsin ( \frac{3}{5} )}{2} + π 1)) = \frac{1}{2}

Refine

0.5 = 0.5

\Rightarrow True

Check the solution

\frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + πn :

False

\frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + πn

Plug in

n = 1

\frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + π 1

For

sec (2 x) + tan (2 x) = \frac{1}{2}

plug in

x = \frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + π 1

sec (2 (\frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + π 1)) + tan (2 (\frac{π}{2} + \frac{arcsin ( \frac{3}{5} )}{2} + π 1)) = \frac{1}{2}

Refine

- 0.5 = 0.5

\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

sec (2 x) + tan (2 x) = \frac{1}{2}

plug in

x = \frac{3 π}{4} + π 1

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

Undefined

\Rightarrow False

x = - \frac{arcsin ( \frac{3}{5} )}{2} + πn

Show solutions in decimal form

x = - \frac{0.64350 \dots}{2} + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sec(2x)+tan(2x)= 1/2 ?
The general solution for sec(2x)+tan(2x)= 1/2 is x=-(0.64350…)/2+pin