What is the general solution for 3sin(t)=-3+cos(t) ?

The general solution for 3sin(t)=-3+cos(t) is t=(3pi)/2+2pin,t=2pi-0.92729…+2pin

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Popular >

3sin(t)=-3+cos(t)

Solution

3 sin (t) = - 3 + cos (t)

Solution

t = \frac{3 π}{2} + 2 πn, t = 2 π - 0.92729 \dots + 2 πn

Degrees

t = 27 0^{\circ} + 36 0^{\circ} n, t = 306.86989 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (t) = - 3 + cos (t)

Square both sides

(3 sin (t))^{2} = (- 3 + cos (t))^{2}

Subtract

(- 3 + cos (t))^{2}

from both sides

9 sin^{2} (t) - 9 + 6 cos (t) - cos^{2} (t) = 0

Rewrite using trig identities

- 9 - cos^{2} (t) + 6 cos (t) + 9 sin^{2} (t)

Use the Pythagorean identity:

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

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

= - 9 - cos^{2} (t) + 6 cos (t) + 9 (1 - cos^{2} (t))

Simplify

- 9 - cos^{2} (t) + 6 cos (t) + 9 (1 - cos^{2} (t)) : 6 cos (t) - 10 cos^{2} (t)

- 9 - cos^{2} (t) + 6 cos (t) + 9 (1 - cos^{2} (t))

Expand

9 (1 - cos^{2} (t)) : 9 - 9 cos^{2} (t)

9 (1 - cos^{2} (t))

Apply the distributive law:

a (b - c) = ab - a c

a = 9, b = 1, c = cos^{2} (t)

= 9 \cdot 1 - 9 cos^{2} (t)

Multiply the numbers:

9 \cdot 1 = 9

= 9 - 9 cos^{2} (t)

= - 9 - cos^{2} (t) + 6 cos (t) + 9 - 9 cos^{2} (t)

Simplify

- 9 - cos^{2} (t) + 6 cos (t) + 9 - 9 cos^{2} (t) : 6 cos (t) - 10 cos^{2} (t)

- 9 - cos^{2} (t) + 6 cos (t) + 9 - 9 cos^{2} (t)

Group like terms

= - cos^{2} (t) + 6 cos (t) - 9 cos^{2} (t) - 9 + 9

Add similar elements:

- cos^{2} (t) - 9 cos^{2} (t) = - 10 cos^{2} (t)

= - 10 cos^{2} (t) + 6 cos (t) - 9 + 9

- 9 + 9 = 0

= 6 cos (t) - 10 cos^{2} (t)

= 6 cos (t) - 10 cos^{2} (t)

= 6 cos (t) - 10 cos^{2} (t)

- 10 cos^{2} (t) + 6 cos (t) = 0

Solve by substitution

- 10 cos^{2} (t) + 6 cos (t) = 0

Let:

cos (t) = u

- 10 u^{2} + 6 u = 0

- 10 u^{2} + 6 u = 0 : u = 0, u = \frac{3}{5}

- 10 u^{2} + 6 u = 0

Solve with the quadratic formula

- 10 u^{2} + 6 u = 0

Quadratic Equation Formula:

For

a = - 10, b = 6, c = 0

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 10 ) \cdot 0}{2 ( - 10 )}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 10 ) \cdot 0}{2 ( - 10 )}

6^{2} - 4 (- 10) \cdot 0 = 6

6^{2} - 4 (- 10) \cdot 0

Apply rule

- (- a) = a

= 6^{2} + 4 \cdot 10 \cdot 0

Apply rule

0 \cdot a = 0

= 6^{2} + 0

6^{2} + 0 = 6^{2}

= 6^{2}

Apply radical rule:

assuming

a \geq 0

= 6

u_{1, 2} = \frac{- 6 \pm 6}{2 ( - 10 )}

Separate the solutions

u_{1} = \frac{- 6 + 6}{2 ( - 10 )}, u_{2} = \frac{- 6 - 6}{2 ( - 10 )}

u = \frac{- 6 + 6}{2 ( - 10 )} : 0

\frac{- 6 + 6}{2 ( - 10 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 + 6}{- 2 \cdot 10}

Add/Subtract the numbers:

- 6 + 6 = 0

= \frac{0}{- 2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{0}{- 20}

Apply the fraction rule:

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

= - \frac{0}{20}

Apply rule

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

= - 0

= 0

u = \frac{- 6 - 6}{2 ( - 10 )} : \frac{3}{5}

\frac{- 6 - 6}{2 ( - 10 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 - 6}{- 2 \cdot 10}

Subtract the numbers:

- 6 - 6 = - 12

= \frac{- 12}{- 2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 12}{- 20}

Apply the fraction rule:

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

= \frac{12}{20}

Cancel the common factor:

4

= \frac{3}{5}

The solutions to the quadratic equation are:

u = 0, u = \frac{3}{5}

Substitute back

u = cos (t)

cos (t) = 0, cos (t) = \frac{3}{5}

cos (t) = 0, cos (t) = \frac{3}{5}

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

cos (t) = \frac{3}{5}

Apply trig inverse properties

cos (t) = \frac{3}{5}

General solutions for

cos (t) = \frac{3}{5}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

t = arccos (\frac{3}{5}) + 2 πn, t = 2 π - arccos (\frac{3}{5}) + 2 πn

t = arccos (\frac{3}{5}) + 2 πn, t = 2 π - arccos (\frac{3}{5}) + 2 πn

Combine all the solutions

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn, t = arccos (\frac{3}{5}) + 2 πn, t = 2 π - arccos (\frac{3}{5}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

3 sin (t) = - 3 + cos (t)

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

Check the solution

\frac{π}{2} + 2 πn :

False

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

3 sin (t) = - 3 + cos (t)

plug in

t = \frac{π}{2} + 2 π 1

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

Refine

3 = - 3

\Rightarrow False

Check the solution

\frac{3 π}{2} + 2 πn :

True

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

3 sin (t) = - 3 + cos (t)

plug in

t = \frac{3 π}{2} + 2 π 1

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

Refine

- 3 = - 3

\Rightarrow True

Check the solution

arccos (\frac{3}{5}) + 2 πn :

False

arccos (\frac{3}{5}) + 2 πn

Plug in

n = 1

arccos (\frac{3}{5}) + 2 π 1

For

3 sin (t) = - 3 + cos (t)

plug in

t = arccos (\frac{3}{5}) + 2 π 1

3 sin (arccos (\frac{3}{5}) + 2 π 1) = - 3 + cos (arccos (\frac{3}{5}) + 2 π 1)

Refine

2.4 = - 2.4

\Rightarrow False

Check the solution

2 π - arccos (\frac{3}{5}) + 2 πn :

True

2 π - arccos (\frac{3}{5}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{3}{5}) + 2 π 1

For

3 sin (t) = - 3 + cos (t)

plug in

t = 2 π - arccos (\frac{3}{5}) + 2 π 1

3 sin (2 π - arccos (\frac{3}{5}) + 2 π 1) = - 3 + cos (2 π - arccos (\frac{3}{5}) + 2 π 1)

Refine

- 2.4 = - 2.4

\Rightarrow True

t = \frac{3 π}{2} + 2 πn, t = 2 π - arccos (\frac{3}{5}) + 2 πn

Show solutions in decimal form

t = \frac{3 π}{2} + 2 πn, t = 2 π - 0.92729 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3sin(t)=-3+cos(t) ?
The general solution for 3sin(t)=-3+cos(t) is t=(3pi)/2+2pin,t=2pi-0.92729…+2pin