What is the general solution for 2sec(2x)+tan(2x)-3=0 ?

The general solution for 2sec(2x)+tan(2x)-3=0 is x=(0.56432…)/2+pin,x=-(1.20782…)/2+pin

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2sec(2x)+tan(2x)-3=0

Solution

2 sec (2 x) + tan (2 x) - 3 = 0

Solution

x = \frac{0.56432 \dots}{2} + πn, x = - \frac{1.20782 \dots}{2} + πn

Degrees

x = 16.16676 \dots^{\circ} + 18 0^{\circ} n, x = - 34.60171 \dots^{\circ} + 18 0^{\circ} n

Solution steps

2 sec (2 x) + tan (2 x) - 3 = 0

Express with sin, cos

2 \cdot \frac{1}{cos ( 2 x )} + \frac{sin ( 2 x )}{cos ( 2 x )} - 3 = 0

Simplify

2 \cdot \frac{1}{cos ( 2 x )} + \frac{sin ( 2 x )}{cos ( 2 x )} - 3 : \frac{2 + sin ( 2 x ) - 3 cos ( 2 x )}{cos ( 2 x )}

2 \cdot \frac{1}{cos ( 2 x )} + \frac{sin ( 2 x )}{cos ( 2 x )} - 3

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 )}

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

Combine the fractions

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

Apply rule

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

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

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

Convert element to fraction:

3 = \frac{3 cos ( 2 x )}{cos ( 2 x )}

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

\frac{2 + sin ( 2 x ) - 3 cos ( 2 x )}{cos ( 2 x )} = 0

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

2 + sin (2 x) - 3 cos (2 x) = 0

Add

3 cos (2 x)

to both sides

2 + sin (2 x) = 3 cos (2 x)

Square both sides

(2 + sin (2 x))^{2} = (3 cos (2 x))^{2}

Subtract

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

from both sides

(2 + sin (2 x))^{2} - 9 cos^{2} (2 x) = 0

Rewrite using trig identities

(2 + sin (2 x))^{2} - 9 cos^{2} (2 x)

Use the Pythagorean identity:

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

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

= (2 + sin (2 x))^{2} - 9 (1 - sin^{2} (2 x))

Simplify

(2 + sin (2 x))^{2} - 9 (1 - sin^{2} (2 x)) : 10 sin^{2} (2 x) + 4 sin (2 x) - 5

(2 + sin (2 x))^{2} - 9 (1 - sin^{2} (2 x))

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

Apply Perfect Square Formula:

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

a = 2, b = sin (2 x)

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

Simplify

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

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

2^{2} = 4

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

Multiply the numbers:

2 \cdot 2 = 4

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

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

= 4 + 4 sin (2 x) + sin^{2} (2 x) - 9 (1 - sin^{2} (2 x))

Expand

- 9 (1 - sin^{2} (2 x)) : - 9 + 9 sin^{2} (2 x)

- 9 (1 - sin^{2} (2 x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 9, b = 1, c = sin^{2} (2 x)

= - 9 \cdot 1 - (- 9) sin^{2} (2 x)

Apply minus-plus rules

- (- a) = a

= - 9 \cdot 1 + 9 sin^{2} (2 x)

Multiply the numbers:

9 \cdot 1 = 9

= - 9 + 9 sin^{2} (2 x)

= 4 + 4 sin (2 x) + sin^{2} (2 x) - 9 + 9 sin^{2} (2 x)

Simplify

4 + 4 sin (2 x) + sin^{2} (2 x) - 9 + 9 sin^{2} (2 x) : 10 sin^{2} (2 x) + 4 sin (2 x) - 5

4 + 4 sin (2 x) + sin^{2} (2 x) - 9 + 9 sin^{2} (2 x)

Group like terms

= 4 sin (2 x) + sin^{2} (2 x) + 9 sin^{2} (2 x) + 4 - 9

Add similar elements:

sin^{2} (2 x) + 9 sin^{2} (2 x) = 10 sin^{2} (2 x)

= 4 sin (2 x) + 10 sin^{2} (2 x) + 4 - 9

Add/Subtract the numbers:

4 - 9 = - 5

= 10 sin^{2} (2 x) + 4 sin (2 x) - 5

= 10 sin^{2} (2 x) + 4 sin (2 x) - 5

= 10 sin^{2} (2 x) + 4 sin (2 x) - 5

- 5 + 10 sin^{2} (2 x) + 4 sin (2 x) = 0

Solve by substitution

- 5 + 10 sin^{2} (2 x) + 4 sin (2 x) = 0

Let:

sin (2 x) = u

- 5 + 10 u^{2} + 4 u = 0

- 5 + 10 u^{2} + 4 u = 0 : u = \frac{- 2 + 3 6}{10}, u = - \frac{2 + 3 6}{10}

- 5 + 10 u^{2} + 4 u = 0

Write in the standard form

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

10 u^{2} + 4 u - 5 = 0

Solve with the quadratic formula

10 u^{2} + 4 u - 5 = 0

Quadratic Equation Formula:

For

a = 10, b = 4, c = - 5

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 10 ( - 5 )}{2 \cdot 10}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 10 ( - 5 )}{2 \cdot 10}

4^{2} - 4 \cdot 10 (- 5) = 66

4^{2} - 4 \cdot 10 (- 5)

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 10 \cdot 5

Multiply the numbers:

4 \cdot 10 \cdot 5 = 200

= 4^{2} + 200

4^{2} = 16

= 16 + 200

Add the numbers:

16 + 200 = 216

= 216

Prime factorization of

216 : 2^{3} \cdot 3^{3}

216

216

divides by

2216 = 108 \cdot 2

= 2 \cdot 108

108

divides by

2108 = 54 \cdot 2

= 2 \cdot 2 \cdot 54

54

divides by

254 = 27 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 27

27

divides by

327 = 9 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 9

9

divides by

39 = 3 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3

= 2^{3} \cdot 3^{3}

= 2^{3} \cdot 3^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 3^{2} \cdot 2 \cdot 3

Apply radical rule:

= 2^{2} 3^{2} 2 \cdot 3

Apply radical rule:

2^{2} = 2

= 2 3^{2} 2 \cdot 3

Apply radical rule:

3^{2} = 3

= 2 \cdot 3 2 \cdot 3

Refine

= 66

u_{1, 2} = \frac{- 4 \pm 6 6}{2 \cdot 10}

Separate the solutions

u_{1} = \frac{- 4 + 6 6}{2 \cdot 10}, u_{2} = \frac{- 4 - 6 6}{2 \cdot 10}

u = \frac{- 4 + 6 6}{2 \cdot 10} : \frac{- 2 + 3 6}{10}

\frac{- 4 + 6 6}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 4 + 6 6}{20}

Factor

- 4 + 66 : 2 (- 2 + 36)

- 4 + 66

Rewrite as

= - 2 \cdot 2 + 2 \cdot 36

Factor out common term

2

= 2 (- 2 + 36)

= \frac{2 ( - 2 + 3 6 )}{20}

Cancel the common factor:

2

= \frac{- 2 + 3 6}{10}

u = \frac{- 4 - 6 6}{2 \cdot 10} : - \frac{2 + 3 6}{10}

\frac{- 4 - 6 6}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 4 - 6 6}{20}

Factor

- 4 - 66 : - 2 (2 + 36)

- 4 - 66

Rewrite as

= - 2 \cdot 2 - 2 \cdot 36

Factor out common term

2

= - 2 (2 + 36)

= - \frac{2 ( 2 + 3 6 )}{20}

Cancel the common factor:

2

= - \frac{2 + 3 6}{10}

The solutions to the quadratic equation are:

u = \frac{- 2 + 3 6}{10}, u = - \frac{2 + 3 6}{10}

Substitute back

u = sin (2 x)

sin (2 x) = \frac{- 2 + 3 6}{10}, sin (2 x) = - \frac{2 + 3 6}{10}

sin (2 x) = \frac{- 2 + 3 6}{10}, sin (2 x) = - \frac{2 + 3 6}{10}

sin (2 x) = \frac{- 2 + 3 6}{10} : x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

sin (2 x) = \frac{- 2 + 3 6}{10}

Apply trig inverse properties

sin (2 x) = \frac{- 2 + 3 6}{10}

General solutions for

sin (2 x) = \frac{- 2 + 3 6}{10}

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

2 x = arcsin (\frac{- 2 + 3 6}{10}) + 2 πn, 2 x = π - arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

2 x = arcsin (\frac{- 2 + 3 6}{10}) + 2 πn, 2 x = π - arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

Solve

2 x = arcsin (\frac{- 2 + 3 6}{10}) + 2 πn : x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

2 x = arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

Divide both sides by

2

2 x = arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

Solve

2 x = π - arcsin (\frac{- 2 + 3 6}{10}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

2 x = π - arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

Divide both sides by

2

2 x = π - arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

x = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

sin (2 x) = - \frac{2 + 3 6}{10} : x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

sin (2 x) = - \frac{2 + 3 6}{10}

Apply trig inverse properties

sin (2 x) = - \frac{2 + 3 6}{10}

General solutions for

sin (2 x) = - \frac{2 + 3 6}{10}

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

2 x = arcsin (- \frac{2 + 3 6}{10}) + 2 πn, 2 x = π + arcsin (\frac{2 + 3 6}{10}) + 2 πn

2 x = arcsin (- \frac{2 + 3 6}{10}) + 2 πn, 2 x = π + arcsin (\frac{2 + 3 6}{10}) + 2 πn

Solve

2 x = arcsin (- \frac{2 + 3 6}{10}) + 2 πn : x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

2 x = arcsin (- \frac{2 + 3 6}{10}) + 2 πn

Simplify

arcsin (- \frac{2 + 3 6}{10}) + 2 πn : - arcsin (\frac{2 + 3 6}{10}) + 2 πn

arcsin (- \frac{2 + 3 6}{10}) + 2 πn

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{2 + 3 6}{10}) = - arcsin (\frac{2 + 3 6}{10})

= - arcsin (\frac{2 + 3 6}{10}) + 2 πn

2 x = - arcsin (\frac{2 + 3 6}{10}) + 2 πn

Divide both sides by

2

2 x = - arcsin (\frac{2 + 3 6}{10}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + \frac{2 πn}{2}

Simplify

x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

Solve

2 x = π + arcsin (\frac{2 + 3 6}{10}) + 2 πn : x = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

2 x = π + arcsin (\frac{2 + 3 6}{10}) + 2 πn

Divide both sides by

2

2 x = π + arcsin (\frac{2 + 3 6}{10}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

x = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

Combine all the solutions

x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn, x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

2 sec (2 x) + tan (2 x) - 3 = 0

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

Check the solution

\frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn :

True

\frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

Plug in

n = 1

\frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1

For

2 sec (2 x) + tan (2 x) - 3 = 0

plug in

x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1

2 sec 2 \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1 + tan 2 \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1 - 3 = 0

Refine

0 = 0

\Rightarrow True

Check the solution

\frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn :

False

\frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

Plug in

n = 1

\frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1

For

2 sec (2 x) + tan (2 x) - 3 = 0

plug in

x = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1

2 sec 2 \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1 + tan 2 \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1 - 3 = 0

Refine

- 6 = 0

\Rightarrow False

Check the solution

- \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn :

True

- \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

Plug in

n = 1

- \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1

For

2 sec (2 x) + tan (2 x) - 3 = 0

plug in

x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1

2 sec 2 - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1 + tan 2 - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1 - 3 = 0

Refine

0 = 0

\Rightarrow True

Check the solution

\frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn :

False

\frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

Plug in

n = 1

\frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1

For

2 sec (2 x) + tan (2 x) - 3 = 0

plug in

x = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1

2 sec 2 \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1 + tan 2 \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1 - 3 = 0

Refine

- 6 = 0

\Rightarrow False

x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn, x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

Show solutions in decimal form

x = \frac{0.56432 \dots}{2} + πn, x = - \frac{1.20782 \dots}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2sec(2x)+tan(2x)-3=0 ?
The general solution for 2sec(2x)+tan(2x)-3=0 is x=(0.56432…)/2+pin,x=-(1.20782…)/2+pin