What is the general solution for cos^2(t)-sin^2(t)=-1 ?

The general solution for cos^2(t)-sin^2(t)=-1 is t= pi/2+pin

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cos^2(t)-sin^2(t)=-1

cos^{2} (t) - sin^{2} (t) = - 1

t = \frac{π}{2} + πn

Degrees

t = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

cos^{2} (t) - sin^{2} (t) = - 1

Rewrite using trig identities

cos^{2} (t) - sin^{2} (t)

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

= cos (2 t)

cos (2 t) = - 1

General solutions for

cos (2 t) = - 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}

2 t = π + 2 πn

2 t = π + 2 πn

Solve

2 t = π + 2 πn : t = \frac{π}{2} + πn

2 t = π + 2 πn

Divide both sides by

2

2 t = π + 2 πn

Divide both sides by

2

\frac{2 t}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

t = \frac{π}{2} + πn

t = \frac{π}{2} + πn

t = \frac{π}{2} + πn

- a^{2} c_{1} cos (a x) - a^{2} c_{2} sin (a x) = 0

10 = - 12 sin (x) + 1.8 cos (x)

8 tan (B) + 12 = 3 tan (B) + 7

- 2 + sin (2 θ) = - \frac{3}{2}

tan (c) = \frac{24}{7}

What is the general solution for cos^2(t)-sin^2(t)=-1 ?
The general solution for cos^2(t)-sin^2(t)=-1 is t= pi/2+pin