What is the general solution for cos^4(t)=1 ?

The general solution for cos^4(t)=1 is t=2pin,t=pi+2pin

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Popular >

cos^4(t)=1

Solution

cos^{4} (t) = 1

Solution

t = 2 πn, t = π + 2 πn

Degrees

t = 0^{\circ} + 36 0^{\circ} n, t = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{4} (t) = 1

Solve by substitution

cos^{4} (t) = 1

Let:

cos (t) = u

u^{4} = 1

u^{4} = 1 : u = 1, u = - 1, u = i, u = - i

u^{4} = 1

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} = 1

Solve

v^{2} = 1 : v = 1, v = - 1

v^{2} = 1

For

(g (x))^{2} = f (a)

the solutions are

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

v = 1, v = - 1

v = 1, v = - 1

Substitute back

v = u^{2},

solve for

u

Solve

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

u^{2} = 1

Apply rule

1 = 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

Solve

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

u^{2} = - 1

Apply rule

1 = 1

u^{2} = - 1

For

x^{2} = f (a)

the solutions are

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

u = - 1, u = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

The solutions are

u = 1, u = - 1, u = i, u = - i

Substitute back

u = cos (t)

cos (t) = 1, cos (t) = - 1, cos (t) = i, cos (t) = - i

cos (t) = 1, cos (t) = - 1, cos (t) = i, cos (t) = - i

cos (t) = 1 : t = 2 πn

cos (t) = 1

General solutions for

cos (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}

t = 0 + 2 πn

t = 0 + 2 πn

Solve

t = 0 + 2 πn : t = 2 πn

t = 0 + 2 πn

0 + 2 πn = 2 πn

t = 2 πn

t = 2 πn

cos (t) = - 1 : t = π + 2 πn

cos (t) = - 1

General solutions for

cos (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}

t = π + 2 πn

t = π + 2 πn

cos (t) = i :

No Solution

cos (t) = i

No Solution

cos (t) = - i :

No Solution

cos (t) = - i

No Solution

Combine all the solutions

t = 2 πn, t = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos^4(t)=1 ?
The general solution for cos^4(t)=1 is t=2pin,t=pi+2pin