What is the general solution for -6sin(t)+6cos(2t)=0 ?

The general solution for -6sin(t)+6cos(2t)=0 is t=(3pi)/2+2pin,t= pi/6+2pin,t=(5pi)/6+2pin

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-6sin(t)+6cos(2t)=0

Solution

- 6 sin (t) + 6 cos (2 t) = 0

Solution

t = \frac{3 π}{2} + 2 πn, t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

Degrees

t = 27 0^{\circ} + 36 0^{\circ} n, t = 3 0^{\circ} + 36 0^{\circ} n, t = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

- 6 sin (t) + 6 cos (2 t) = 0

Rewrite using trig identities

6 cos (2 t) - 6 sin (t)

Use the Double Angle identity:

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

= 6 (1 - 2 sin^{2} (t)) - 6 sin (t)

(1 - 2 sin^{2} (t)) \cdot 6 - 6 sin (t) = 0

Solve by substitution

(1 - 2 sin^{2} (t)) \cdot 6 - 6 sin (t) = 0

Let:

sin (t) = u

(1 - 2 u^{2}) \cdot 6 - 6 u = 0

(1 - 2 u^{2}) \cdot 6 - 6 u = 0 : u = - 1, u = \frac{1}{2}

(1 - 2 u^{2}) \cdot 6 - 6 u = 0

Expand

(1 - 2 u^{2}) \cdot 6 - 6 u : 6 - 12 u^{2} - 6 u

(1 - 2 u^{2}) \cdot 6 - 6 u

= 6 (1 - 2 u^{2}) - 6 u

Expand

6 (1 - 2 u^{2}) : 6 - 12 u^{2}

6 (1 - 2 u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 6, b = 1, c = 2 u^{2}

= 6 \cdot 1 - 6 \cdot 2 u^{2}

Simplify

6 \cdot 1 - 6 \cdot 2 u^{2} : 6 - 12 u^{2}

6 \cdot 1 - 6 \cdot 2 u^{2}

Multiply the numbers:

6 \cdot 1 = 6

= 6 - 6 \cdot 2 u^{2}

Multiply the numbers:

6 \cdot 2 = 12

= 6 - 12 u^{2}

= 6 - 12 u^{2}

= 6 - 12 u^{2} - 6 u

6 - 12 u^{2} - 6 u = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 12, b = - 6, c = 6

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 12 ) \cdot 6}{2 ( - 12 )}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 12 ) \cdot 6}{2 ( - 12 )}

(- 6)^{2} - 4 (- 12) \cdot 6 = 18

(- 6)^{2} - 4 (- 12) \cdot 6

Apply rule

- (- a) = a

= (- 6)^{2} + 4 \cdot 12 \cdot 6

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} + 4 \cdot 12 \cdot 6

Multiply the numbers:

4 \cdot 12 \cdot 6 = 288

= 6^{2} + 288

6^{2} = 36

= 36 + 288

Add the numbers:

36 + 288 = 324

= 324

Factor the number:

324 = 1 8^{2}

= 1 8^{2}

Apply radical rule:

1 8^{2} = 18

= 18

u_{1, 2} = \frac{- ( - 6 ) \pm 18}{2 ( - 12 )}

Separate the solutions

u_{1} = \frac{- ( - 6 ) + 18}{2 ( - 12 )}, u_{2} = \frac{- ( - 6 ) - 18}{2 ( - 12 )}

u = \frac{- ( - 6 ) + 18}{2 ( - 12 )} : - 1

\frac{- ( - 6 ) + 18}{2 ( - 12 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{6 + 18}{- 2 \cdot 12}

Add the numbers:

6 + 18 = 24

= \frac{24}{- 2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{24}{- 24}

Apply the fraction rule:

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

= - \frac{24}{24}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 6 ) - 18}{2 ( - 12 )} : \frac{1}{2}

\frac{- ( - 6 ) - 18}{2 ( - 12 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{6 - 18}{- 2 \cdot 12}

Subtract the numbers:

6 - 18 = - 12

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

Multiply the numbers:

2 \cdot 12 = 24

= \frac{- 12}{- 24}

Apply the fraction rule:

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

= \frac{12}{24}

Cancel the common factor:

12

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (t)

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

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

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

sin (t) = - 1

General solutions for

sin (t) = - 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

t = \frac{3 π}{2} + 2 πn

t = \frac{3 π}{2} + 2 πn

sin (t) = \frac{1}{2} : t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

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

General solutions for

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

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

Combine all the solutions

t = \frac{3 π}{2} + 2 πn, t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -6sin(t)+6cos(2t)=0 ?
The general solution for -6sin(t)+6cos(2t)=0 is t=(3pi)/2+2pin,t= pi/6+2pin,t=(5pi)/6+2pin