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

The general solution for -1sec^2(θ)+2sec(θ)=1 is θ=2pin

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Popular >

-1sec^2(θ)+2sec(θ)=1

Solution

- 1 sec^{2} (θ) + 2 sec (θ) = 1

Solution

θ = 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

- 1 \cdot sec^{2} (θ) + 2 sec (θ) = 1

Solve by substitution

- 1 \cdot sec^{2} (θ) + 2 sec (θ) = 1

Let:

sec (θ) = u

- 1 \cdot u^{2} + 2 u = 1

- 1 \cdot u^{2} + 2 u = 1 : u = 1

- 1 \cdot u^{2} + 2 u = 1

Refine

- u^{2} + 2 u = 1

Move

1

to the left side

- u^{2} + 2 u = 1

Subtract

1

from both sides

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

Simplify

- u^{2} + 2 u - 1 = 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 = sec (θ)

sec (θ) = 1

sec (θ) = 1

sec (θ) = 1 : θ = 2 πn

sec (θ) = 1

General solutions for

sec (θ) = 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}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Combine all the solutions

θ = 2 πn

Popular Examples

cos(θ)=0.8701

cos (θ) = 0.8701

cos(3v)=cos(6),0<= v<= 360

cos (3 v) = cos (6), 0^{\circ} \leq v^{\circ} \leq 36 0^{\circ}

tan(x)=(25.644)/(106.372)

tan (x) = \frac{25.644}{106.372}

sin^2(θ)-7sin(θ)=0

sin^{2} (θ) - 7 sin (θ) = 0

-1=2cos(2θ)

- 1 = 2 cos (2 θ)

Frequently Asked Questions (FAQ)

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