What is the general solution for 4sin(θ)=4cos(2θ) ?

The general solution for 4sin(θ)=4cos(2θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin

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4sin(θ)=4cos(2θ)

Solution

4 sin (θ) = 4 cos (2 θ)

Solution

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

Degrees

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

Solution steps

4 sin (θ) = 4 cos (2 θ)

Subtract

4 cos (2 θ)

from both sides

4 sin (θ) - 4 cos (2 θ) = 0

Rewrite using trig identities

- 4 cos (2 θ) + 4 sin (θ)

Use the Double Angle identity:

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

= - 4 (1 - 2 sin^{2} (θ)) + 4 sin (θ)

- (1 - 2 sin^{2} (θ)) \cdot 4 + 4 sin (θ) = 0

Solve by substitution

- (1 - 2 sin^{2} (θ)) \cdot 4 + 4 sin (θ) = 0

Let:

sin (θ) = u

- (1 - 2 u^{2}) \cdot 4 + 4 u = 0

- (1 - 2 u^{2}) \cdot 4 + 4 u = 0 : u = \frac{1}{2}, u = - 1

- (1 - 2 u^{2}) \cdot 4 + 4 u = 0

Expand

- (1 - 2 u^{2}) \cdot 4 + 4 u : - 4 + 8 u^{2} + 4 u

- (1 - 2 u^{2}) \cdot 4 + 4 u

= - 4 (1 - 2 u^{2}) + 4 u

Expand

- 4 (1 - 2 u^{2}) : - 4 + 8 u^{2}

- 4 (1 - 2 u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = 2 u^{2}

= - 4 \cdot 1 - (- 4) \cdot 2 u^{2}

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 \cdot 2 u^{2}

Simplify

- 4 \cdot 1 + 4 \cdot 2 u^{2} : - 4 + 8 u^{2}

- 4 \cdot 1 + 4 \cdot 2 u^{2}

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 \cdot 2 u^{2}

Multiply the numbers:

4 \cdot 2 = 8

= - 4 + 8 u^{2}

= - 4 + 8 u^{2}

= - 4 + 8 u^{2} + 4 u

- 4 + 8 u^{2} + 4 u = 0

Write in the standard form

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

8 u^{2} + 4 u - 4 = 0

Solve with the quadratic formula

8 u^{2} + 4 u - 4 = 0

Quadratic Equation Formula:

For

a = 8, b = 4, c = - 4

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

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

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

4^{2} - 4 \cdot 8 (- 4)

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 8 \cdot 4

Multiply the numbers:

4 \cdot 8 \cdot 4 = 128

= 4^{2} + 128

4^{2} = 16

= 16 + 128

Add the numbers:

16 + 128 = 144

= 144

Factor the number:

144 = 1 2^{2}

= 1 2^{2}

Apply radical rule:

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- 4 \pm 12}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- 4 + 12}{2 \cdot 8}, u_{2} = \frac{- 4 - 12}{2 \cdot 8}

u = \frac{- 4 + 12}{2 \cdot 8} : \frac{1}{2}

\frac{- 4 + 12}{2 \cdot 8}

Add/Subtract the numbers:

- 4 + 12 = 8

= \frac{8}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{8}{16}

Cancel the common factor:

8

= \frac{1}{2}

u = \frac{- 4 - 12}{2 \cdot 8} : - 1

\frac{- 4 - 12}{2 \cdot 8}

Subtract the numbers:

- 4 - 12 = - 16

= \frac{- 16}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 16}{16}

Apply the fraction rule:

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

= - \frac{16}{16}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{2}, sin (θ) = - 1

sin (θ) = \frac{1}{2}, sin (θ) = - 1

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

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

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin(θ)=4cos(2θ) ?
The general solution for 4sin(θ)=4cos(2θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin