What is the general solution for 3sec^2(x)-2tan(x)-4=0 ?

The general solution for 3sec^2(x)-2tan(x)-4=0 is x= pi/4+pin,x=-0.32175…+pin

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3sec^2(x)-2tan(x)-4=0

Solution

3 sec^{2} (x) - 2 tan (x) - 4 = 0

Solution

x = \frac{π}{4} + πn, x = - 0.32175 \dots + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = - 18.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

3 sec^{2} (x) - 2 tan (x) - 4 = 0

Rewrite using trig identities

- 4 - 2 tan (x) + 3 sec^{2} (x)

Use the Pythagorean identity:

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

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

Simplify

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

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

Expand

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

3 (tan^{2} (x) + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = 3, b = tan^{2} (x), c = 1

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

Multiply the numbers:

3 \cdot 1 = 3

= 3 tan^{2} (x) + 3

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

Simplify

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

- 4 - 2 tan (x) + 3 tan^{2} (x) + 3

Group like terms

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

Add/Subtract the numbers:

- 4 + 3 = - 1

= 3 tan^{2} (x) - 2 tan (x) - 1

= 3 tan^{2} (x) - 2 tan (x) - 1

= 3 tan^{2} (x) - 2 tan (x) - 1

- 1 - 2 tan (x) + 3 tan^{2} (x) = 0

Solve by substitution

- 1 - 2 tan (x) + 3 tan^{2} (x) = 0

Let:

tan (x) = u

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

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

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

Write in the standard form

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

3 u^{2} - 2 u - 1 = 0

Solve with the quadratic formula

3 u^{2} - 2 u - 1 = 0

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 3 \cdot 1 = 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 3}

Separate the solutions

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

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

\frac{- ( - 2 ) + 4}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{2 + 4}{2 \cdot 3}

Add the numbers:

2 + 4 = 6

= \frac{6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= 1

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

\frac{- ( - 2 ) - 4}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{2 - 4}{2 \cdot 3}

Subtract the numbers:

2 - 4 = - 2

= \frac{- 2}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2}{6}

Apply the fraction rule:

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

= - \frac{2}{6}

Cancel the common factor:

2

= - \frac{1}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = tan (x)

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

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

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

tan (x) = - \frac{1}{3} : x = arctan (- \frac{1}{3}) + πn

tan (x) = - \frac{1}{3}

Apply trig inverse properties

tan (x) = - \frac{1}{3}

General solutions for

tan (x) = - \frac{1}{3}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- \frac{1}{3}) + πn

x = arctan (- \frac{1}{3}) + πn

Combine all the solutions

x = \frac{π}{4} + πn, x = arctan (- \frac{1}{3}) + πn

Show solutions in decimal form

x = \frac{π}{4} + πn, x = - 0.32175 \dots + πn

Popular Examples

9sin^2(x)-1=0

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

sin(x)= 1/(3.5)

sin (x) = \frac{1}{3.5}

tan(θ)=5,96

tan (θ) = 5, 96

1=1-cos(θ)

1 = 1 - cos (θ)

tan(θ/3)=3sin(48)

tan (\frac{θ}{3}) = 3 sin (4 8^{\circ})

Frequently Asked Questions (FAQ)

What is the general solution for 3sec^2(x)-2tan(x)-4=0 ?
The general solution for 3sec^2(x)-2tan(x)-4=0 is x= pi/4+pin,x=-0.32175…+pin