What is the general solution for tan^2(θ)=2sec(θ)-1 ?

The general solution for tan^2(θ)=2sec(θ)-1 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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tan^2(θ)=2sec(θ)-1

Solution

tan^{2} (θ) = 2 sec (θ) - 1

Solution

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Degrees

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

tan^{2} (θ) = 2 sec (θ) - 1

Subtract

2 sec (θ) - 1

from both sides

tan^{2} (θ) - 2 sec (θ) + 1 = 0

Rewrite using trig identities

1 + tan^{2} (θ) - 2 sec (θ)

Use the Pythagorean identity:

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

= - 2 sec (θ) + sec^{2} (θ)

sec^{2} (θ) - 2 sec (θ) = 0

Solve by substitution

sec^{2} (θ) - 2 sec (θ) = 0

Let:

sec (θ) = u

u^{2} - 2 u = 0

u^{2} - 2 u = 0 : u = 2, u = 0

u^{2} - 2 u = 0

Solve with the quadratic formula

u^{2} - 2 u = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot 0 = 2

(- 2)^{2} - 4 \cdot 1 \cdot 0

Apply exponent rule:

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

n

is even

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

= 2^{2} - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a,

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

2 + 2 = 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

u = \frac{- ( - 2 ) - 2}{2 \cdot 1} : 0

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

Apply rule

- (- a) = a

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

Subtract the numbers:

2 - 2 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 2, u = 0

Substitute back

u = sec (θ)

sec (θ) = 2, sec (θ) = 0

sec (θ) = 2, sec (θ) = 0

sec (θ) = 2 : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

sec (θ) = 2

General solutions for

sec (θ) = 2

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}

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

sec (θ) = 0 :

No Solution

sec (θ) = 0

sec (x) \leq - 1 or sec (x) \geq 1

No Solution

Combine all the solutions

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)=2sec(θ)-1 ?
The general solution for tan^2(θ)=2sec(θ)-1 is θ= pi/3+2pin,θ=(5pi)/3+2pin