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

The general solution for 2tan(θ)=sec^2(θ) is θ= pi/4+pin

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2tan(θ)=sec^2(θ)

Solution

2 tan (θ) = sec^{2} (θ)

Solution

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

2 tan (θ) = sec^{2} (θ)

Subtract

sec^{2} (θ)

from both sides

2 tan (θ) - sec^{2} (θ) = 0

Rewrite using trig identities

- sec^{2} (θ) + 2 tan (θ)

Use the Pythagorean identity:

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

= - (tan^{2} (θ) + 1) + 2 tan (θ)

- (tan^{2} (θ) + 1) : - tan^{2} (θ) - 1

- (tan^{2} (θ) + 1)

Distribute parentheses

= - (tan^{2} (θ)) - (1)

Apply minus-plus rules

+ (- a) = - a

= - tan^{2} (θ) - 1

= - tan^{2} (θ) - 1 + 2 tan (θ)

- 1 - tan^{2} (θ) + 2 tan (θ) = 0

Solve by substitution

- 1 - tan^{2} (θ) + 2 tan (θ) = 0

Let:

tan (θ) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

2^{2} - 4 (- 1) (- 1)

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

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

u = \frac{- 2}{2 ( - 1 )}

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

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

Remove parentheses:

(- a) = - a

= \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

u = 1

The solution to the quadratic equation is:

u = 1

Substitute back

u = tan (θ)

tan (θ) = 1

tan (θ) = 1

tan (θ) = 1 : θ = \frac{π}{4} + πn

tan (θ) = 1

General solutions for

tan (θ) = 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}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Combine all the solutions

θ = \frac{π}{4} + πn

Popular Examples

tan(θ)= 14/15

tan (θ) = \frac{14}{15}

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

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

sin(-θ)=-3/5 ,tan(θ)>0

sin (- θ) = - \frac{3}{5}, tan (θ) > 0

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

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

tan^2(x)+5=3sec^2(x)

tan^{2} (x) + 5 = 3 sec^{2} (x)

Frequently Asked Questions (FAQ)

What is the general solution for 2tan(θ)=sec^2(θ) ?
The general solution for 2tan(θ)=sec^2(θ) is θ= pi/4+pin