What is the general solution for sec^2(θ)-1=0,0<= θ<= 2pi ?

Q: What is the general solution for sec^2(θ)-1=0,0<= θ<= 2pi ?

The general solution for sec^2(θ)-1=0,0<= θ<= 2pi is θ=0,θ=2pi,θ=pi

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sec^2(θ)-1=0,0<= θ<= 2pi

Solution

sec^{2} (θ) - 1 = 0, 0 \leq θ \leq 2 π

Solution

θ = 0, θ = 2 π, θ = π

Degrees

θ = 0^{\circ}, θ = 36 0^{\circ}, θ = 18 0^{\circ}

Solution steps

sec^{2} (θ) - 1 = 0, 0 \leq θ \leq 2 π

Solve by substitution

sec^{2} (θ) - 1 = 0

Let:

sec (θ) = u

u^{2} - 1 = 0

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

u^{2} - 1 = 0

Move

1

to the right side

u^{2} - 1 = 0

Add

1

to both sides

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

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = sec (θ)

sec (θ) = 1, sec (θ) = - 1

sec (θ) = 1, sec (θ) = - 1

sec (θ) = 1, 0 \leq θ \leq 2 π : θ = 0, θ = 2 π

sec (θ) = 1, 0 \leq θ \leq 2 π

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

Solutions for the range

0 \leq θ \leq 2 π

θ = 0, θ = 2 π

sec (θ) = - 1, 0 \leq θ \leq 2 π : θ = π

sec (θ) = - 1, 0 \leq θ \leq 2 π

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 = π + 2 πn

θ = π + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = π

Combine all the solutions

θ = 0, θ = 2 π, θ = π

Popular Examples

2cos^2(θ)-3cos(θ)=0,(3pi)/2 <θ<2pi

2 cos^{2} (θ) - 3 cos (θ) = 0, \frac{3 π}{2} < θ < 2 π

cos(θ)= 9/17

cos (θ) = \frac{9}{17}

-1=(tan(x)-(3/4))/(1+tan(x)(3/4))

- 1 = \frac{tan ( x ) - ( \frac{3}{4} )}{1 + tan ( x ) ( \frac{3}{4} )}

cos(x)=-15/13

cos (x) = - \frac{15}{13}

arctan((20)/x)-arctan((400)/x)=-9.72

arctan (\frac{20}{x}) - arctan (\frac{400}{x}) = - 9.72

Frequently Asked Questions (FAQ)

What is the general solution for sec^2(θ)-1=0,0<= θ<= 2pi ?
The general solution for sec^2(θ)-1=0,0<= θ<= 2pi is θ=0,θ=2pi,θ=pi