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Popular >

tan^2(3y)-2sec(3y)-2=0

Solution

tan^{2} (3 y) - 2 sec (3 y) - 2 = 0

Solution

y = \frac{1.23095 \dots}{3} + \frac{2 πn}{3}, y = \frac{2 π}{3} - \frac{1.23095 \dots}{3} + \frac{2 πn}{3}, y = \frac{π}{3} + \frac{2 πn}{3}

Degrees

y = 23.50959 \dots^{\circ} + 12 0^{\circ} n, y = 96.49040 \dots^{\circ} + 12 0^{\circ} n, y = 6 0^{\circ} + 12 0^{\circ} n

Solution steps

tan^{2} (3 y) - 2 sec (3 y) - 2 = 0

Rewrite using trig identities

- 2 + tan^{2} (3 y) - 2 sec (3 y)

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= - 2 + sec^{2} (3 y) - 1 - 2 sec (3 y)

Simplify

- 2 + sec^{2} (3 y) - 1 - 2 sec (3 y) : sec^{2} (3 y) - 2 sec (3 y) - 3

- 2 + sec^{2} (3 y) - 1 - 2 sec (3 y)

Group like terms

= sec^{2} (3 y) - 2 sec (3 y) - 2 - 1

Subtract the numbers:

- 2 - 1 = - 3

= sec^{2} (3 y) - 2 sec (3 y) - 3

= sec^{2} (3 y) - 2 sec (3 y) - 3

- 3 + sec^{2} (3 y) - 2 sec (3 y) = 0

Solve by substitution

- 3 + sec^{2} (3 y) - 2 sec (3 y) = 0

Let:

sec (3 y) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = - 2, c = - 3

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

Add the numbers:

4 + 12 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 4}{2 \cdot 1}, u_{2} = \frac{- ( - 2 ) - 4}{2 \cdot 1}

u = \frac{- ( - 2 ) + 4}{2 \cdot 1} : 3

\frac{- ( - 2 ) + 4}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 + 4}{2 \cdot 1}

Add the numbers:

2 + 4 = 6

= \frac{6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

u = \frac{- ( - 2 ) - 4}{2 \cdot 1} : - 1

\frac{- ( - 2 ) - 4}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 - 4}{2 \cdot 1}

Subtract the numbers:

2 - 4 = - 2

= \frac{- 2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 3, u = - 1

Substitute back

u = sec (3 y)

sec (3 y) = 3, sec (3 y) = - 1

sec (3 y) = 3, sec (3 y) = - 1

sec (3 y) = 3 : y = \frac{arcsec ( 3 )}{3} + \frac{2 πn}{3}, y = \frac{2 π}{3} - \frac{arcsec ( 3 )}{3} + \frac{2 πn}{3}

sec (3 y) = 3

Apply trig inverse properties

sec (3 y) = 3

General solutions for

sec (3 y) = 3

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = 2 π - arcsec (a) + 2 πn

3 y = arcsec (3) + 2 πn, 3 y = 2 π - arcsec (3) + 2 πn

3 y = arcsec (3) + 2 πn, 3 y = 2 π - arcsec (3) + 2 πn

Solve

3 y = arcsec (3) + 2 πn : y = \frac{arcsec ( 3 )}{3} + \frac{2 πn}{3}

3 y = arcsec (3) + 2 πn

Divide both sides by

3

3 y = arcsec (3) + 2 πn

Divide both sides by

3

\frac{3 y}{3} = \frac{arcsec ( 3 )}{3} + \frac{2 πn}{3}

Simplify

y = \frac{arcsec ( 3 )}{3} + \frac{2 πn}{3}

y = \frac{arcsec ( 3 )}{3} + \frac{2 πn}{3}

Solve

3 y = 2 π - arcsec (3) + 2 πn : y = \frac{2 π}{3} - \frac{arcsec ( 3 )}{3} + \frac{2 πn}{3}

3 y = 2 π - arcsec (3) + 2 πn

Divide both sides by

3

3 y = 2 π - arcsec (3) + 2 πn

Divide both sides by

3

\frac{3 y}{3} = \frac{2 π}{3} - \frac{arcsec ( 3 )}{3} + \frac{2 πn}{3}

Simplify

y = \frac{2 π}{3} - \frac{arcsec ( 3 )}{3} + \frac{2 πn}{3}

y = \frac{2 π}{3} - \frac{arcsec ( 3 )}{3} + \frac{2 πn}{3}

y = \frac{arcsec ( 3 )}{3} + \frac{2 πn}{3}, y = \frac{2 π}{3} - \frac{arcsec ( 3 )}{3} + \frac{2 πn}{3}

sec (3 y) = - 1 : y = \frac{π}{3} + \frac{2 πn}{3}

sec (3 y) = - 1

General solutions for

sec (3 y) = - 1

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

3 y = π + 2 πn

3 y = π + 2 πn

Solve

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

3 y = π + 2 πn

Divide both sides by

3

3 y = π + 2 πn

Divide both sides by

3

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

Simplify

y = \frac{π}{3} + \frac{2 πn}{3}

y = \frac{π}{3} + \frac{2 πn}{3}

y = \frac{π}{3} + \frac{2 πn}{3}

Combine all the solutions

y = \frac{arcsec ( 3 )}{3} + \frac{2 πn}{3}, y = \frac{2 π}{3} - \frac{arcsec ( 3 )}{3} + \frac{2 πn}{3}, y = \frac{π}{3} + \frac{2 πn}{3}

Show solutions in decimal form

y = \frac{1.23095 \dots}{3} + \frac{2 πn}{3}, y = \frac{2 π}{3} - \frac{1.23095 \dots}{3} + \frac{2 πn}{3}, y = \frac{π}{3} + \frac{2 πn}{3}

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