What is the general solution for 3cos(2θ)=3sin(θ) ?

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

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3cos(2θ)=3sin(θ)

Solution

3 cos (2 θ) = 3 sin (θ)

Solution

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

Degrees

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

Solution steps

3 cos (2 θ) = 3 sin (θ)

Subtract

3 sin (θ)

from both sides

3 cos (2 θ) - 3 sin (θ) = 0

Rewrite using trig identities

3 cos (2 θ) - 3 sin (θ)

Use the Double Angle identity:

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

= 3 (1 - 2 sin^{2} (θ)) - 3 sin (θ)

(1 - 2 sin^{2} (θ)) \cdot 3 - 3 sin (θ) = 0

Solve by substitution

(1 - 2 sin^{2} (θ)) \cdot 3 - 3 sin (θ) = 0

Let:

sin (θ) = u

(1 - 2 u^{2}) \cdot 3 - 3 u = 0

(1 - 2 u^{2}) \cdot 3 - 3 u = 0 : u = - 1, u = \frac{1}{2}

(1 - 2 u^{2}) \cdot 3 - 3 u = 0

Expand

(1 - 2 u^{2}) \cdot 3 - 3 u : 3 - 6 u^{2} - 3 u

(1 - 2 u^{2}) \cdot 3 - 3 u

= 3 (1 - 2 u^{2}) - 3 u

Expand

3 (1 - 2 u^{2}) : 3 - 6 u^{2}

3 (1 - 2 u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = 2 u^{2}

= 3 \cdot 1 - 3 \cdot 2 u^{2}

Simplify

3 \cdot 1 - 3 \cdot 2 u^{2} : 3 - 6 u^{2}

3 \cdot 1 - 3 \cdot 2 u^{2}

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 \cdot 2 u^{2}

Multiply the numbers:

3 \cdot 2 = 6

= 3 - 6 u^{2}

= 3 - 6 u^{2}

= 3 - 6 u^{2} - 3 u

3 - 6 u^{2} - 3 u = 0

Write in the standard form

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

- 6 u^{2} - 3 u + 3 = 0

Solve with the quadratic formula

- 6 u^{2} - 3 u + 3 = 0

Quadratic Equation Formula:

For

a = - 6, b = - 3, c = 3

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 6 ) \cdot 3}{2 ( - 6 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 6 ) \cdot 3}{2 ( - 6 )}

(- 3)^{2} - 4 (- 6) \cdot 3 = 9

(- 3)^{2} - 4 (- 6) \cdot 3

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 6 \cdot 3

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 6 \cdot 3

Multiply the numbers:

4 \cdot 6 \cdot 3 = 72

= 3^{2} + 72

3^{2} = 9

= 9 + 72

Add the numbers:

9 + 72 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 3 ) \pm 9}{2 ( - 6 )}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 9}{2 ( - 6 )}, u_{2} = \frac{- ( - 3 ) - 9}{2 ( - 6 )}

u = \frac{- ( - 3 ) + 9}{2 ( - 6 )} : - 1

\frac{- ( - 3 ) + 9}{2 ( - 6 )}

Remove parentheses:

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

= \frac{3 + 9}{- 2 \cdot 6}

Add the numbers:

3 + 9 = 12

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{12}{- 12}

Apply the fraction rule:

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

= - \frac{12}{12}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 3 ) - 9}{2 ( - 6 )} : \frac{1}{2}

\frac{- ( - 3 ) - 9}{2 ( - 6 )}

Remove parentheses:

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

= \frac{3 - 9}{- 2 \cdot 6}

Subtract the numbers:

3 - 9 = - 6

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6}{- 12}

Apply the fraction rule:

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

= \frac{6}{12}

Cancel the common factor:

6

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

sin (θ) = - 1, sin (θ) = \frac{1}{2}

sin (θ) = - 1, sin (θ) = \frac{1}{2}

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

sin (θ) = - 1

General solutions for

sin (θ) = - 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}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

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

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \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}

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

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

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

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