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

The general solution for tan(x)-sec(x)=3,0<= x<2pi is x=pi+0.92729…

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Popular Trigonometry >

tan(x)-sec(x)=3,0<= x<2pi

Solution

tan (x) - sec (x) = 3, 0 \leq x < 2 π

Solution

x = π + 0.92729 \dots

Degrees

x = 233.13010 \dots^{\circ}

Solution steps

tan (x) - sec (x) = 3, 0 \leq x < 2 π

Subtract

3

from both sides

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

Express with sin, cos

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

Simplify

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

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

Combine the fractions

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

Apply rule

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

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

= \frac{sin ( x ) - 1}{cos ( x )} - 3

Convert element to fraction:

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

= \frac{sin ( x ) - 1}{cos ( x )} - \frac{3 cos ( x )}{cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( x ) - 1 - 3 cos ( x )}{cos ( x )}

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

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

sin (x) - 1 - 3 cos (x) = 0

Add

3 cos (x)

to both sides

sin (x) - 1 = 3 cos (x)

Square both sides

(sin (x) - 1)^{2} = (3 cos (x))^{2}

Subtract

(3 cos (x))^{2}

from both sides

(sin (x) - 1)^{2} - 9 cos^{2} (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

(- 1 + sin (x))^{2} - 9 (1 - sin^{2} (x)) : 10 sin^{2} (x) - 2 sin (x) - 8

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

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

Apply Perfect Square Formula:

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

a = - 1, b = sin (x)

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

Simplify

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

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

Remove parentheses:

(- a) = - a

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

(- 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

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

2 \cdot 1 \cdot sin (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 sin (x)

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

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

9 \cdot 1 = 9

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

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

Simplify

1 - 2 sin (x) + sin^{2} (x) - 9 + 9 sin^{2} (x) : 10 sin^{2} (x) - 2 sin (x) - 8

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

Group like terms

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

Add similar elements:

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

= - 2 sin (x) + 10 sin^{2} (x) + 1 - 9

Add/Subtract the numbers:

1 - 9 = - 8

= 10 sin^{2} (x) - 2 sin (x) - 8

= 10 sin^{2} (x) - 2 sin (x) - 8

= 10 sin^{2} (x) - 2 sin (x) - 8

- 8 + 10 sin^{2} (x) - 2 sin (x) = 0

Solve by substitution

- 8 + 10 sin^{2} (x) - 2 sin (x) = 0

Let:

sin (x) = u

- 8 + 10 u^{2} - 2 u = 0

- 8 + 10 u^{2} - 2 u = 0 : u = 1, u = - \frac{4}{5}

- 8 + 10 u^{2} - 2 u = 0

Write in the standard form

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

10 u^{2} - 2 u - 8 = 0

Solve with the quadratic formula

10 u^{2} - 2 u - 8 = 0

Quadratic Equation Formula:

For

a = 10, b = - 2, c = - 8

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 10 ( - 8 )}{2 \cdot 10}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 10 ( - 8 )}{2 \cdot 10}

(- 2)^{2} - 4 \cdot 10 (- 8) = 18

(- 2)^{2} - 4 \cdot 10 (- 8)

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 10 \cdot 8

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 10 \cdot 8

Multiply the numbers:

4 \cdot 10 \cdot 8 = 320

= 2^{2} + 320

2^{2} = 4

= 4 + 320

Add the numbers:

4 + 320 = 324

= 324

Factor the number:

324 = 1 8^{2}

= 1 8^{2}

Apply radical rule:

1 8^{2} = 18

= 18

u_{1, 2} = \frac{- ( - 2 ) \pm 18}{2 \cdot 10}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 18}{2 \cdot 10}, u_{2} = \frac{- ( - 2 ) - 18}{2 \cdot 10}

u = \frac{- ( - 2 ) + 18}{2 \cdot 10} : 1

\frac{- ( - 2 ) + 18}{2 \cdot 10}

Apply rule

- (- a) = a

= \frac{2 + 18}{2 \cdot 10}

Add the numbers:

2 + 18 = 20

= \frac{20}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{20}{20}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 18}{2 \cdot 10} : - \frac{4}{5}

\frac{- ( - 2 ) - 18}{2 \cdot 10}

Apply rule

- (- a) = a

= \frac{2 - 18}{2 \cdot 10}

Subtract the numbers:

2 - 18 = - 16

= \frac{- 16}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 16}{20}

Apply the fraction rule:

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

= - \frac{16}{20}

Cancel the common factor:

4

= - \frac{4}{5}

The solutions to the quadratic equation are:

u = 1, u = - \frac{4}{5}

Substitute back

u = sin (x)

sin (x) = 1, sin (x) = - \frac{4}{5}

sin (x) = 1, sin (x) = - \frac{4}{5}

sin (x) = 1, 0 \leq x < 2 π : x = \frac{π}{2}

sin (x) = 1, 0 \leq x < 2 π

General solutions for

sin (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}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{2}

sin (x) = - \frac{4}{5}, 0 \leq x < 2 π : x = π + arcsin (\frac{4}{5}), x = - arcsin (\frac{4}{5}) + 2 π

sin (x) = - \frac{4}{5}, 0 \leq x < 2 π

Apply trig inverse properties

sin (x) = - \frac{4}{5}

General solutions for

sin (x) = - \frac{4}{5}

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

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

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

Solutions for the range

0 \leq x < 2 π

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

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

tan (x) - sec (x) = 3

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

Check the solution

\frac{π}{2} :

False

\frac{π}{2}

Plug in

n = 1

\frac{π}{2}

For

tan (x) - sec (x) = 3

plug in

x = \frac{π}{2}

tan (\frac{π}{2}) - sec (\frac{π}{2}) = 3

Undefined

\Rightarrow False

Check the solution

π + arcsin (\frac{4}{5}) :

True

π + arcsin (\frac{4}{5})

Plug in

n = 1

π + arcsin (\frac{4}{5})

For

tan (x) - sec (x) = 3

plug in

x = π + arcsin (\frac{4}{5})

tan (π + arcsin (\frac{4}{5})) - sec (π + arcsin (\frac{4}{5})) = 3

Refine

3 = 3

\Rightarrow True

Check the solution

- arcsin (\frac{4}{5}) + 2 π :

False

- arcsin (\frac{4}{5}) + 2 π

Plug in

n = 1

- arcsin (\frac{4}{5}) + 2 π

For

tan (x) - sec (x) = 3

plug in

x = - arcsin (\frac{4}{5}) + 2 π

tan (- arcsin (\frac{4}{5}) + 2 π) - sec (- arcsin (\frac{4}{5}) + 2 π) = 3

Refine

- 3 = 3

\Rightarrow False

x = π + arcsin (\frac{4}{5})

Show solutions in decimal form

x = π + 0.92729 \dots

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for tan(x)-sec(x)=3,0<= x<2pi ?
The general solution for tan(x)-sec(x)=3,0<= x<2pi is x=pi+0.92729…