What is the general solution for 8-32cos^2(t)=0 ?

The general solution for 8-32cos^2(t)=0 is t= pi/3+2pin,t=(5pi)/3+2pin,t=(2pi)/3+2pin,t=(4pi)/3+2pin

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8-32cos^2(t)=0

Solution

8 - 32 cos^{2} (t) = 0

Solution

t = \frac{π}{3} + 2 πn, t = \frac{5 π}{3} + 2 πn, t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

Degrees

t = 6 0^{\circ} + 36 0^{\circ} n, t = 30 0^{\circ} + 36 0^{\circ} n, t = 12 0^{\circ} + 36 0^{\circ} n, t = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

8 - 32 cos^{2} (t) = 0

Solve by substitution

8 - 32 cos^{2} (t) = 0

Let:

cos (t) = u

8 - 32 u^{2} = 0

8 - 32 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

8 - 32 u^{2} = 0

Move

8

to the right side

8 - 32 u^{2} = 0

Subtract

8

from both sides

8 - 32 u^{2} - 8 = 0 - 8

Simplify

- 32 u^{2} = - 8

- 32 u^{2} = - 8

Divide both sides by

- 32

- 32 u^{2} = - 8

Divide both sides by

- 32

\frac{- 32 u ^{2}}{- 32} = \frac{- 8}{- 32}

Simplify

\frac{- 32 u ^{2}}{- 32} = \frac{- 8}{- 32}

Simplify

\frac{- 32 u ^{2}}{- 32} : u^{2}

\frac{- 32 u ^{2}}{- 32}

Apply the fraction rule:

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

= \frac{32 u ^{2}}{32}

Divide the numbers:

\frac{32}{32} = 1

= u^{2}

Simplify

\frac{- 8}{- 32} : \frac{1}{4}

\frac{- 8}{- 32}

Apply the fraction rule:

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

= \frac{8}{32}

Cancel the common factor:

8

= \frac{1}{4}

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

For

x^{2} = f (a)

the solutions are

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

u = \frac{1}{4}, u = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

- \frac{1}{4}

Simplify

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2}

u = \frac{1}{2}, u = - \frac{1}{2}

Substitute back

u = cos (t)

cos (t) = \frac{1}{2}, cos (t) = - \frac{1}{2}

cos (t) = \frac{1}{2}, cos (t) = - \frac{1}{2}

cos (t) = \frac{1}{2} : t = \frac{π}{3} + 2 πn, t = \frac{5 π}{3} + 2 πn

cos (t) = \frac{1}{2}

General solutions for

cos (t) = \frac{1}{2}

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 = \frac{π}{3} + 2 πn, t = \frac{5 π}{3} + 2 πn

t = \frac{π}{3} + 2 πn, t = \frac{5 π}{3} + 2 πn

cos (t) = - \frac{1}{2} : t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

cos (t) = - \frac{1}{2}

General solutions for

cos (t) = - \frac{1}{2}

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 = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

Combine all the solutions

t = \frac{π}{3} + 2 πn, t = \frac{5 π}{3} + 2 πn, t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 8-32cos^2(t)=0 ?
The general solution for 8-32cos^2(t)=0 is t= pi/3+2pin,t=(5pi)/3+2pin,t=(2pi)/3+2pin,t=(4pi)/3+2pin