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

The general solution for tan^2(θ)-sec^2(θ)=cos(-θ) is θ=-pi-2pin

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tan^2(θ)-sec^2(θ)=cos(-θ)

tan^{2} (θ) - sec^{2} (θ) = cos (- θ)

θ = - π - 2 πn

Degrees

θ = - 18 0^{\circ} - 36 0^{\circ} n

Solution steps

tan^{2} (θ) - sec^{2} (θ) = cos (- θ)

Rewrite using trig identities

tan^{2} (θ) - sec^{2} (θ)

Use the Pythagorean identity:

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

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

= - 1

- 1 = cos (- θ)

Switch sides

cos (- θ) = - 1

General solutions for

cos (- θ) = - 1

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

- θ = π + 2 πn

- θ = π + 2 πn

Solve

- θ = π + 2 πn : θ = - π - 2 πn

- θ = π + 2 πn

Divide both sides by

- 1

- θ = π + 2 πn

Divide both sides by

- 1

\frac{- θ}{- 1} = \frac{π}{- 1} + \frac{2 πn}{- 1}

Simplify

\frac{- θ}{- 1} = \frac{π}{- 1} + \frac{2 πn}{- 1}

Simplify

\frac{- θ}{- 1} : θ

\frac{- θ}{- 1}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{θ}{1}

Apply rule

\frac{a}{1} = a

= θ

Simplify

\frac{π}{- 1} + \frac{2 πn}{- 1} : - π - 2 πn

\frac{π}{- 1} + \frac{2 πn}{- 1}

\frac{π}{- 1} = - π

\frac{π}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{π}{1}

Apply rule

\frac{a}{1} = a

= - π

= - π + \frac{2 πn}{- 1}

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - π - 2 πn

θ = - π - 2 πn

θ = - π - 2 πn

θ = - π - 2 πn

θ = - π - 2 πn

- 2 sin (t) - 4 cos (2 t) = 0

8 sin (2 x) - 43 = 0

cos (3 x) cos (9 x) + sin (3 x) sin (9 x) = 0

cos (2 x) + 5 sin (x) + 2 = 0, 0 \leq x < 2 π

8 - 12 sin^{2} (θ) = 4 cos^{2} (θ)

What is the general solution for tan^2(θ)-sec^2(θ)=cos(-θ) ?
The general solution for tan^2(θ)-sec^2(θ)=cos(-θ) is θ=-pi-2pin