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

The general solution for sec(θ)=sin^2(θ)sec(θ) is No Solution for θ\in\mathbb{R}

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Popular >

sec(θ)=sin^2(θ)sec(θ)

Solution

sec (θ) = sin^{2} (θ) sec (θ)

Solution

No Solution for θ \in R

Solution steps

sec (θ) = sin^{2} (θ) sec (θ)

Subtract

sin^{2} (θ) sec (θ)

from both sides

sec (θ) - sin^{2} (θ) sec (θ) = 0

Factor

sec (θ) - sin^{2} (θ) sec (θ) : - sec (θ) (sin (θ) + 1) (sin (θ) - 1)

sec (θ) - sin^{2} (θ) sec (θ)

Factor out common term

- sec (θ)

= - sec (θ) (- 1 + sin^{2} (θ))

Factor

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

sin^{2} (θ) - 1

Rewrite

1

1^{2}

= sin^{2} (θ) - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

sin^{2} (θ) - 1^{2} = (sin (θ) + 1) (sin (θ) - 1)

= (sin (θ) + 1) (sin (θ) - 1)

= - sec (θ) (sin (θ) + 1) (sin (θ) - 1)

- sec (θ) (sin (θ) + 1) (sin (θ) - 1) = 0

Solving each part separately

sec (θ) = 0 or sin (θ) + 1 = 0 or sin (θ) - 1 = 0

sec (θ) = 0 :

No Solution

sec (θ) = 0

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

No Solution

sin (θ) + 1 = 0 : θ = \frac{3 π}{2} + 2 πn

sin (θ) + 1 = 0

Move

1

to the right side

sin (θ) + 1 = 0

Subtract

1

from both sides

sin (θ) + 1 - 1 = 0 - 1

Simplify

sin (θ) = - 1

sin (θ) = - 1

General solutions for

sin (θ) = - 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

sin (θ) - 1 = 0 : θ = \frac{π}{2} + 2 πn

sin (θ) - 1 = 0

Move

1

to the right side

sin (θ) - 1 = 0

Add

1

to both sides

sin (θ) - 1 + 1 = 0 + 1

Simplify

sin (θ) = 1

sin (θ) = 1

General solutions for

sin (θ) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Combine all the solutions

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

Since the equation is undefined for:

\frac{3 π}{2} + 2 πn, \frac{π}{2} + 2 πn

No Solution for θ \in R

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2 tan (θ) + 2 = 0

Frequently Asked Questions (FAQ)

What is the general solution for sec(θ)=sin^2(θ)sec(θ) ?
The general solution for sec(θ)=sin^2(θ)sec(θ) is No Solution for θ\in\mathbb{R}