What is the general solution for 8sin(t)+8cos(t)=8 ?

The general solution for 8sin(t)+8cos(t)=8 is t= pi/2+2pin,t=2pin

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Popular Trigonometry >

8sin(t)+8cos(t)=8

Solution

8 sin (t) + 8 cos (t) = 8

Solution

t = \frac{π}{2} + 2 πn, t = 2 πn

Degrees

t = 9 0^{\circ} + 36 0^{\circ} n, t = 0^{\circ} + 36 0^{\circ} n

Solution steps

8 sin (t) + 8 cos (t) = 8

Subtract

8 cos (t)

from both sides

8 sin (t) = 8 - 8 cos (t)

Square both sides

(8 sin (t))^{2} = (8 - 8 cos (t))^{2}

Subtract

(8 - 8 cos (t))^{2}

from both sides

64 sin^{2} (t) - 64 + 128 cos (t) - 64 cos^{2} (t) = 0

Rewrite using trig identities

- 64 + 128 cos (t) - 64 cos^{2} (t) + 64 sin^{2} (t)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

= - 64 + 128 cos (t) - 64 cos^{2} (t) + 64 (1 - cos^{2} (t))

Simplify

- 64 + 128 cos (t) - 64 cos^{2} (t) + 64 (1 - cos^{2} (t)) : 128 cos (t) - 128 cos^{2} (t)

- 64 + 128 cos (t) - 64 cos^{2} (t) + 64 (1 - cos^{2} (t))

Expand

64 (1 - cos^{2} (t)) : 64 - 64 cos^{2} (t)

64 (1 - cos^{2} (t))

Apply the distributive law:

a (b - c) = ab - a c

a = 64, b = 1, c = cos^{2} (t)

= 64 \cdot 1 - 64 cos^{2} (t)

Multiply the numbers:

64 \cdot 1 = 64

= 64 - 64 cos^{2} (t)

= - 64 + 128 cos (t) - 64 cos^{2} (t) + 64 - 64 cos^{2} (t)

Simplify

- 64 + 128 cos (t) - 64 cos^{2} (t) + 64 - 64 cos^{2} (t) : 128 cos (t) - 128 cos^{2} (t)

- 64 + 128 cos (t) - 64 cos^{2} (t) + 64 - 64 cos^{2} (t)

Group like terms

= 128 cos (t) - 64 cos^{2} (t) - 64 cos^{2} (t) - 64 + 64

Add similar elements:

- 64 cos^{2} (t) - 64 cos^{2} (t) = - 128 cos^{2} (t)

= 128 cos (t) - 128 cos^{2} (t) - 64 + 64

- 64 + 64 = 0

= 128 cos (t) - 128 cos^{2} (t)

= 128 cos (t) - 128 cos^{2} (t)

= 128 cos (t) - 128 cos^{2} (t)

128 cos (t) - 128 cos^{2} (t) = 0

Solve by substitution

128 cos (t) - 128 cos^{2} (t) = 0

Let:

cos (t) = u

128 u - 128 u^{2} = 0

128 u - 128 u^{2} = 0 : u = 0, u = 1

128 u - 128 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 128 u^{2} + 128 u = 0

Solve with the quadratic formula

- 128 u^{2} + 128 u = 0

Quadratic Equation Formula:

For

a = - 128, b = 128, c = 0

u_{1, 2} = \frac{- 128 \pm 12 8 ^{2} - 4 ( - 128 ) \cdot 0}{2 ( - 128 )}

u_{1, 2} = \frac{- 128 \pm 12 8 ^{2} - 4 ( - 128 ) \cdot 0}{2 ( - 128 )}

12 8^{2} - 4 (- 128) \cdot 0 = 128

12 8^{2} - 4 (- 128) \cdot 0

Apply rule

- (- a) = a

= 12 8^{2} + 4 \cdot 128 \cdot 0

Apply rule

0 \cdot a = 0

= 12 8^{2} + 0

12 8^{2} + 0 = 12 8^{2}

= 12 8^{2}

Apply radical rule:

assuming

a \geq 0

= 128

u_{1, 2} = \frac{- 128 \pm 128}{2 ( - 128 )}

Separate the solutions

u_{1} = \frac{- 128 + 128}{2 ( - 128 )}, u_{2} = \frac{- 128 - 128}{2 ( - 128 )}

u = \frac{- 128 + 128}{2 ( - 128 )} : 0

\frac{- 128 + 128}{2 ( - 128 )}

Remove parentheses:

(- a) = - a

= \frac{- 128 + 128}{- 2 \cdot 128}

Add/Subtract the numbers:

- 128 + 128 = 0

= \frac{0}{- 2 \cdot 128}

Multiply the numbers:

2 \cdot 128 = 256

= \frac{0}{- 256}

Apply the fraction rule:

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

= - \frac{0}{256}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 128 - 128}{2 ( - 128 )} : 1

\frac{- 128 - 128}{2 ( - 128 )}

Remove parentheses:

(- a) = - a

= \frac{- 128 - 128}{- 2 \cdot 128}

Subtract the numbers:

- 128 - 128 = - 256

= \frac{- 256}{- 2 \cdot 128}

Multiply the numbers:

2 \cdot 128 = 256

= \frac{- 256}{- 256}

Apply the fraction rule:

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

= \frac{256}{256}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 0, u = 1

Substitute back

u = cos (t)

cos (t) = 0, cos (t) = 1

cos (t) = 0, cos (t) = 1

cos (t) = 0 : t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

cos (t) = 0

General solutions for

cos (t) = 0

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}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

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

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

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

8 sin (t) + 8 cos (t) = 8

Remove the ones that don't agree with the equation.

Check the solution

\frac{π}{2} + 2 πn :

True

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

8 sin (t) + 8 cos (t) = 8

plug in

t = \frac{π}{2} + 2 π 1

8 sin (\frac{π}{2} + 2 π 1) + 8 cos (\frac{π}{2} + 2 π 1) = 8

Refine

8 = 8

\Rightarrow True

Check the solution

\frac{3 π}{2} + 2 πn :

False

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

8 sin (t) + 8 cos (t) = 8

plug in

t = \frac{3 π}{2} + 2 π 1

8 sin (\frac{3 π}{2} + 2 π 1) + 8 cos (\frac{3 π}{2} + 2 π 1) = 8

Refine

- 8 = 8

\Rightarrow False

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

8 sin (t) + 8 cos (t) = 8

plug in

t = 2 π 1

8 sin (2 π 1) + 8 cos (2 π 1) = 8

Refine

8 = 8

\Rightarrow True

t = \frac{π}{2} + 2 πn, t = 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 8sin(t)+8cos(t)=8 ?
The general solution for 8sin(t)+8cos(t)=8 is t= pi/2+2pin,t=2pin