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

The general solution for 3-3sin(t)=3sqrt(3)cos(t) is t= pi/2+2pin,t=(11pi)/6+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

3 - 3 sin (t) = 33 cos (t)

Solution

t = \frac{π}{2} + 2 πn, t = \frac{11 π}{6} + 2 πn

Degrees

t = 9 0^{\circ} + 36 0^{\circ} n, t = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

3 - 3 sin (t) = 33 cos (t)

Square both sides

(3 - 3 sin (t))^{2} = (33 cos (t))^{2}

Subtract

(33 cos (t))^{2}

from both sides

(3 - 3 sin (t))^{2} - 27 cos^{2} (t) = 0

Rewrite using trig identities

(3 - 3 sin (t))^{2} - 27 cos^{2} (t)

Use the Pythagorean identity:

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

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

= (3 - 3 sin (t))^{2} - 27 (1 - sin^{2} (t))

Simplify

(3 - 3 sin (t))^{2} - 27 (1 - sin^{2} (t)) : 36 sin^{2} (t) - 18 sin (t) - 18

(3 - 3 sin (t))^{2} - 27 (1 - sin^{2} (t))

(3 - 3 sin (t))^{2} : 9 - 18 sin (t) + 9 sin^{2} (t)

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 3, b = 3 sin (t)

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

Simplify

3^{2} - 2 \cdot 3 \cdot 3 sin (t) + (3 sin (t))^{2} : 9 - 18 sin (t) + 9 sin^{2} (t)

3^{2} - 2 \cdot 3 \cdot 3 sin (t) + (3 sin (t))^{2}

3^{2} = 9

3^{2}

3^{2} = 9

= 9

2 \cdot 3 \cdot 3 sin (t) = 18 sin (t)

2 \cdot 3 \cdot 3 sin (t)

Multiply the numbers:

2 \cdot 3 \cdot 3 = 18

= 18 sin (t)

(3 sin (t))^{2} = 9 sin^{2} (t)

(3 sin (t))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 3^{2} sin^{2} (t)

3^{2} = 9

= 9 sin^{2} (t)

= 9 - 18 sin (t) + 9 sin^{2} (t)

= 9 - 18 sin (t) + 9 sin^{2} (t)

= 9 - 18 sin (t) + 9 sin^{2} (t) - 27 (1 - sin^{2} (t))

Expand

- 27 (1 - sin^{2} (t)) : - 27 + 27 sin^{2} (t)

- 27 (1 - sin^{2} (t))

Apply the distributive law:

a (b - c) = ab - a c

a = - 27, b = 1, c = sin^{2} (t)

= - 27 \cdot 1 - (- 27) sin^{2} (t)

Apply minus-plus rules

- (- a) = a

= - 27 \cdot 1 + 27 sin^{2} (t)

Multiply the numbers:

27 \cdot 1 = 27

= - 27 + 27 sin^{2} (t)

= 9 - 18 sin (t) + 9 sin^{2} (t) - 27 + 27 sin^{2} (t)

Simplify

9 - 18 sin (t) + 9 sin^{2} (t) - 27 + 27 sin^{2} (t) : 36 sin^{2} (t) - 18 sin (t) - 18

9 - 18 sin (t) + 9 sin^{2} (t) - 27 + 27 sin^{2} (t)

Group like terms

= - 18 sin (t) + 9 sin^{2} (t) + 27 sin^{2} (t) + 9 - 27

Add similar elements:

9 sin^{2} (t) + 27 sin^{2} (t) = 36 sin^{2} (t)

= - 18 sin (t) + 36 sin^{2} (t) + 9 - 27

Add/Subtract the numbers:

9 - 27 = - 18

= 36 sin^{2} (t) - 18 sin (t) - 18

= 36 sin^{2} (t) - 18 sin (t) - 18

= 36 sin^{2} (t) - 18 sin (t) - 18

- 18 - 18 sin (t) + 36 sin^{2} (t) = 0

Solve by substitution

- 18 - 18 sin (t) + 36 sin^{2} (t) = 0

Let:

sin (t) = u

- 18 - 18 u + 36 u^{2} = 0

- 18 - 18 u + 36 u^{2} = 0 : u = 1, u = - \frac{1}{2}

- 18 - 18 u + 36 u^{2} = 0

Write in the standard form

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

36 u^{2} - 18 u - 18 = 0

Solve with the quadratic formula

36 u^{2} - 18 u - 18 = 0

Quadratic Equation Formula:

For

a = 36, b = - 18, c = - 18

u_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 36 ( - 18 )}{2 \cdot 36}

u_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 36 ( - 18 )}{2 \cdot 36}

(- 18)^{2} - 4 \cdot 36 (- 18) = 54

(- 18)^{2} - 4 \cdot 36 (- 18)

Apply rule

- (- a) = a

= (- 18)^{2} + 4 \cdot 36 \cdot 18

Apply exponent rule:

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

n

is even

(- 18)^{2} = 1 8^{2}

= 1 8^{2} + 4 \cdot 36 \cdot 18

Multiply the numbers:

4 \cdot 36 \cdot 18 = 2592

= 1 8^{2} + 2592

1 8^{2} = 324

= 324 + 2592

Add the numbers:

324 + 2592 = 2916

= 2916

Factor the number:

2916 = 5 4^{2}

= 5 4^{2}

Apply radical rule:

5 4^{2} = 54

= 54

u_{1, 2} = \frac{- ( - 18 ) \pm 54}{2 \cdot 36}

Separate the solutions

u_{1} = \frac{- ( - 18 ) + 54}{2 \cdot 36}, u_{2} = \frac{- ( - 18 ) - 54}{2 \cdot 36}

u = \frac{- ( - 18 ) + 54}{2 \cdot 36} : 1

\frac{- ( - 18 ) + 54}{2 \cdot 36}

Apply rule

- (- a) = a

= \frac{18 + 54}{2 \cdot 36}

Add the numbers:

18 + 54 = 72

= \frac{72}{2 \cdot 36}

Multiply the numbers:

2 \cdot 36 = 72

= \frac{72}{72}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 18 ) - 54}{2 \cdot 36} : - \frac{1}{2}

\frac{- ( - 18 ) - 54}{2 \cdot 36}

Apply rule

- (- a) = a

= \frac{18 - 54}{2 \cdot 36}

Subtract the numbers:

18 - 54 = - 36

= \frac{- 36}{2 \cdot 36}

Multiply the numbers:

2 \cdot 36 = 72

= \frac{- 36}{72}

Apply the fraction rule:

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

= - \frac{36}{72}

Cancel the common factor:

36

= - \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{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{π}{2} + 2 πn

t = \frac{π}{2} + 2 πn

sin (t) = - \frac{1}{2} : t = \frac{7 π}{6} + 2 πn, t = \frac{11 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{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

Combine all the solutions

t = \frac{π}{2} + 2 πn, t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

3 - 3 sin (t) = 33 cos (t)

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

3 - 3 sin (t) = 33 cos (t)

plug in

t = \frac{π}{2} + 2 π 1

3 - 3 sin (\frac{π}{2} + 2 π 1) = 33 cos (\frac{π}{2} + 2 π 1)

Refine

0 = 0

\Rightarrow True

Check the solution

\frac{7 π}{6} + 2 πn :

False

\frac{7 π}{6} + 2 πn

Plug in

n = 1

\frac{7 π}{6} + 2 π 1

For

3 - 3 sin (t) = 33 cos (t)

plug in

t = \frac{7 π}{6} + 2 π 1

3 - 3 sin (\frac{7 π}{6} + 2 π 1) = 33 cos (\frac{7 π}{6} + 2 π 1)

Refine

4.5 = - 4.5

\Rightarrow False

Check the solution

\frac{11 π}{6} + 2 πn :

True

\frac{11 π}{6} + 2 πn

Plug in

n = 1

\frac{11 π}{6} + 2 π 1

For

3 - 3 sin (t) = 33 cos (t)

plug in

t = \frac{11 π}{6} + 2 π 1

3 - 3 sin (\frac{11 π}{6} + 2 π 1) = 33 cos (\frac{11 π}{6} + 2 π 1)

Refine

4.5 = 4.5

\Rightarrow True

t = \frac{π}{2} + 2 πn, t = \frac{11 π}{6} + 2 πn

Popular Examples

sin(θ)= 3/10

sin (θ) = \frac{3}{10}

30arctan(x)=5pi

30 arctan (x) = 5 π

cosh(2x)-10cosh(x)+13=0

cosh (2 x) - 10 cosh (x) + 13 = 0

sqrt(3)sin(x)-sin(2x)=0

3 sin (x) - sin (2 x) = 0

7cos^2(x)+14cos(x)+7=0

7 cos^{2} (x) + 14 cos (x) + 7 = 0

Frequently Asked Questions (FAQ)

What is the general solution for 3-3sin(t)=3sqrt(3)cos(t) ?
The general solution for 3-3sin(t)=3sqrt(3)cos(t) is t= pi/2+2pin,t=(11pi)/6+2pin