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

The general solution for tan(3x)+sec(3x)=2 is x=(0.64350…)/3+(2pin)/3

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

Solution

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

Solution

x = \frac{0.64350 \dots}{3} + \frac{2 πn}{3}

Degrees

x = 12.28996 \dots^{\circ} + 12 0^{\circ} n

Solution steps

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

Subtract

2

from both sides

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

Express with sin, cos

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

Simplify

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

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

Combine the fractions

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

Apply rule

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

sin (3 x) + 1 - 2 cos (3 x) = 0

Add

2 cos (3 x)

to both sides

sin (3 x) + 1 = 2 cos (3 x)

Square both sides

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

Subtract

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

from both sides

(sin (3 x) + 1)^{2} - 4 cos^{2} (3 x) = 0

Rewrite using trig identities

(1 + sin (3 x))^{2} - 4 cos^{2} (3 x)

Use the Pythagorean identity:

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

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

= (1 + sin (3 x))^{2} - 4 (1 - sin^{2} (3 x))

Simplify

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

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

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

Apply Perfect Square Formula:

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

a = 1, b = sin (3 x)

= 1^{2} + 2 \cdot 1 \cdot sin (3 x) + sin^{2} (3 x)

Simplify

1^{2} + 2 \cdot 1 \cdot sin (3 x) + sin^{2} (3 x) : 1 + 2 sin (3 x) + sin^{2} (3 x)

1^{2} + 2 \cdot 1 \cdot sin (3 x) + sin^{2} (3 x)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot sin (3 x) + sin^{2} (3 x)

Multiply the numbers:

2 \cdot 1 = 2

= 1 + 2 sin (3 x) + sin^{2} (3 x)

= 1 + 2 sin (3 x) + sin^{2} (3 x)

= 1 + 2 sin (3 x) + sin^{2} (3 x) - 4 (1 - sin^{2} (3 x))

Expand

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

- 4 (1 - sin^{2} (3 x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (3 x)

= - 4 \cdot 1 - (- 4) sin^{2} (3 x)

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 sin^{2} (3 x)

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 sin^{2} (3 x)

= 1 + 2 sin (3 x) + sin^{2} (3 x) - 4 + 4 sin^{2} (3 x)

Simplify

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

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

Group like terms

= 2 sin (3 x) + sin^{2} (3 x) + 4 sin^{2} (3 x) + 1 - 4

Add similar elements:

sin^{2} (3 x) + 4 sin^{2} (3 x) = 5 sin^{2} (3 x)

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

Add/Subtract the numbers:

1 - 4 = - 3

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

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

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

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

Solve by substitution

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

Let:

sin (3 x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 5, b = 2, c = - 3

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

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

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

2^{2} - 4 \cdot 5 (- 3)

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 5 \cdot 3

Multiply the numbers:

4 \cdot 5 \cdot 3 = 60

= 2^{2} + 60

2^{2} = 4

= 4 + 60

Add the numbers:

4 + 60 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 2 + 8 = 6

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{6}{10}

Cancel the common factor:

2

= \frac{3}{5}

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

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

Subtract the numbers:

- 2 - 8 = - 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 (3 x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Solve

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

3 x = arcsin (\frac{3}{5}) + 2 πn

Divide both sides by

3

3 x = arcsin (\frac{3}{5}) + 2 πn

Divide both sides by

3

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

Simplify

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

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

Solve

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

3 x = π - arcsin (\frac{3}{5}) + 2 πn

Divide both sides by

3

3 x = π - arcsin (\frac{3}{5}) + 2 πn

Divide both sides by

3

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

Simplify

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

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

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

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

sin (3 x) = - 1

General solutions for

sin (3 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}

3 x = \frac{3 π}{2} + 2 πn

3 x = \frac{3 π}{2} + 2 πn

Solve

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

3 x = \frac{3 π}{2} + 2 πn

Divide both sides by

3

3 x = \frac{3 π}{2} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{3 π}{2}}{3} + \frac{2 πn}{3} : \frac{π}{2} + \frac{2 πn}{3}

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

\frac{\frac{3 π}{2}}{3} = \frac{π}{2}

\frac{\frac{3 π}{2}}{3}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{3 π}{6}

Cancel the common factor:

3

= \frac{π}{2}

= \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

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

True

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

Plug in

n = 1

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

For

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

plug in

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

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

Refine

2 = 2

\Rightarrow True

Check the solution

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

False

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

Plug in

n = 1

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

For

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

plug in

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

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

Refine

- 2 = 2

\Rightarrow False

Check the solution

\frac{π}{2} + \frac{2 πn}{3} :

False

\frac{π}{2} + \frac{2 πn}{3}

Plug in

n = 1

\frac{π}{2} + \frac{2 π 1}{3}

For

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

plug in

x = \frac{π}{2} + \frac{2 π 1}{3}

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

Undefined

\Rightarrow False

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

Show solutions in decimal form

x = \frac{0.64350 \dots}{3} + \frac{2 πn}{3}

Popular Examples

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sin(θ)=(-2)/3

sin (θ) = \frac{- 2}{3}

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cos(θ/3)=sin(θ)

cos (\frac{θ}{3}) = sin (θ)

sin(x)-cos(x)= 1/5

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

Frequently Asked Questions (FAQ)

What is the general solution for tan(3x)+sec(3x)=2 ?
The general solution for tan(3x)+sec(3x)=2 is x=(0.64350…)/3+(2pin)/3