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

The general solution for sec^2(x)-4tan(x)=-2 is x=1.24904…+pin,x= pi/4+pin

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

Solution

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

Solution

x = 1.24904 \dots + πn, x = \frac{π}{4} + πn

Degrees

x = 71.56505 \dots^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

- 2

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

Group like terms

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

Add the numbers:

2 + 1 = 3

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

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

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

Solve by substitution

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

Let:

tan (x) = u

3 + u^{2} - 4 u = 0

3 + u^{2} - 4 u = 0 : u = 3, u = 1

3 + u^{2} - 4 u = 0

Write in the standard form

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

u^{2} - 4 u + 3 = 0

Solve with the quadratic formula

u^{2} - 4 u + 3 = 0

Quadratic Equation Formula:

For

a = 1, b = - 4, c = 3

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

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

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

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

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2}

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

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

Subtract the numbers:

16 - 12 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

4 - 2 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 3, u = 1

Substitute back

u = tan (x)

tan (x) = 3, tan (x) = 1

tan (x) = 3, tan (x) = 1

tan (x) = 3 : x = arctan (3) + πn

tan (x) = 3

Apply trig inverse properties

tan (x) = 3

General solutions for

tan (x) = 3

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

x = arctan (3) + πn

x = arctan (3) + πn

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

Combine all the solutions

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

Show solutions in decimal form

x = 1.24904 \dots + πn, x = \frac{π}{4} + πn

Popular Examples

-1=tan(θ)

- 1 = tan (θ)

4cos(x)=2

4 cos (x) = 2

5+2tan(3θ+pi/3)=3

5 + 2 tan (3 θ + \frac{π}{3}) = 3

3sec^2(x)-6=0,0<= x<= 2pi

3 sec^{2} (x) - 6 = 0, 0 \leq x \leq 2 π

cot^2(x)=1

cot^{2} (x) = 1

Frequently Asked Questions (FAQ)

What is the general solution for sec^2(x)-4tan(x)=-2 ?
The general solution for sec^2(x)-4tan(x)=-2 is x=1.24904…+pin,x= pi/4+pin