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

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

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Popular >

4-4sin(θ)=4cos(θ)

Solution

4 - 4 sin (θ) = 4 cos (θ)

Solution

θ = \frac{π}{2} + 2 πn, θ = 2 πn

Degrees

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

Solution steps

4 - 4 sin (θ) = 4 cos (θ)

Square both sides

(4 - 4 sin (θ))^{2} = (4 cos (θ))^{2}

Subtract

(4 cos (θ))^{2}

from both sides

(4 - 4 sin (θ))^{2} - 16 cos^{2} (θ) = 0

Rewrite using trig identities

(4 - 4 sin (θ))^{2} - 16 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= (4 - 4 sin (θ))^{2} - 16 (1 - sin^{2} (θ))

Simplify

(4 - 4 sin (θ))^{2} - 16 (1 - sin^{2} (θ)) : 32 sin^{2} (θ) - 32 sin (θ)

(4 - 4 sin (θ))^{2} - 16 (1 - sin^{2} (θ))

(4 - 4 sin (θ))^{2} : 16 - 32 sin (θ) + 16 sin^{2} (θ)

Apply Perfect Square Formula:

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

a = 4, b = 4 sin (θ)

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

Simplify

4^{2} - 2 \cdot 4 \cdot 4 sin (θ) + (4 sin (θ))^{2} : 16 - 32 sin (θ) + 16 sin^{2} (θ)

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

4^{2} = 16

4^{2}

4^{2} = 16

= 16

2 \cdot 4 \cdot 4 sin (θ) = 32 sin (θ)

2 \cdot 4 \cdot 4 sin (θ)

Multiply the numbers:

2 \cdot 4 \cdot 4 = 32

= 32 sin (θ)

(4 sin (θ))^{2} = 16 sin^{2} (θ)

(4 sin (θ))^{2}

Apply exponent rule:

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

= 4^{2} sin^{2} (θ)

4^{2} = 16

= 16 sin^{2} (θ)

= 16 - 32 sin (θ) + 16 sin^{2} (θ)

= 16 - 32 sin (θ) + 16 sin^{2} (θ)

= 16 - 32 sin (θ) + 16 sin^{2} (θ) - 16 (1 - sin^{2} (θ))

Expand

- 16 (1 - sin^{2} (θ)) : - 16 + 16 sin^{2} (θ)

- 16 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 16, b = 1, c = sin^{2} (θ)

= - 16 \cdot 1 - (- 16) sin^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 16 \cdot 1 + 16 sin^{2} (θ)

Multiply the numbers:

16 \cdot 1 = 16

= - 16 + 16 sin^{2} (θ)

= 16 - 32 sin (θ) + 16 sin^{2} (θ) - 16 + 16 sin^{2} (θ)

Simplify

16 - 32 sin (θ) + 16 sin^{2} (θ) - 16 + 16 sin^{2} (θ) : 32 sin^{2} (θ) - 32 sin (θ)

16 - 32 sin (θ) + 16 sin^{2} (θ) - 16 + 16 sin^{2} (θ)

Group like terms

= - 32 sin (θ) + 16 sin^{2} (θ) + 16 sin^{2} (θ) + 16 - 16

Add similar elements:

16 sin^{2} (θ) + 16 sin^{2} (θ) = 32 sin^{2} (θ)

= - 32 sin (θ) + 32 sin^{2} (θ) + 16 - 16

16 - 16 = 0

= 32 sin^{2} (θ) - 32 sin (θ)

= 32 sin^{2} (θ) - 32 sin (θ)

= 32 sin^{2} (θ) - 32 sin (θ)

- 32 sin (θ) + 32 sin^{2} (θ) = 0

Solve by substitution

- 32 sin (θ) + 32 sin^{2} (θ) = 0

Let:

sin (θ) = u

- 32 u + 32 u^{2} = 0

- 32 u + 32 u^{2} = 0 : u = 1, u = 0

- 32 u + 32 u^{2} = 0

Write in the standard form

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

32 u^{2} - 32 u = 0

Solve with the quadratic formula

32 u^{2} - 32 u = 0

Quadratic Equation Formula:

For

a = 32, b = - 32, c = 0

u_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 32 \cdot 0}{2 \cdot 32}

u_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 32 \cdot 0}{2 \cdot 32}

(- 32)^{2} - 4 \cdot 32 \cdot 0 = 32

(- 32)^{2} - 4 \cdot 32 \cdot 0

Apply exponent rule:

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

n

is even

(- 32)^{2} = 3 2^{2}

= 3 2^{2} - 4 \cdot 32 \cdot 0

Apply rule

0 \cdot a = 0

= 3 2^{2} - 0

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

= 3 2^{2}

Apply radical rule:

assuming

a \geq 0

= 32

u_{1, 2} = \frac{- ( - 32 ) \pm 32}{2 \cdot 32}

Separate the solutions

u_{1} = \frac{- ( - 32 ) + 32}{2 \cdot 32}, u_{2} = \frac{- ( - 32 ) - 32}{2 \cdot 32}

u = \frac{- ( - 32 ) + 32}{2 \cdot 32} : 1

\frac{- ( - 32 ) + 32}{2 \cdot 32}

Apply rule

- (- a) = a

= \frac{32 + 32}{2 \cdot 32}

Add the numbers:

32 + 32 = 64

= \frac{64}{2 \cdot 32}

Multiply the numbers:

2 \cdot 32 = 64

= \frac{64}{64}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 32 ) - 32}{2 \cdot 32} : 0

\frac{- ( - 32 ) - 32}{2 \cdot 32}

Apply rule

- (- a) = a

= \frac{32 - 32}{2 \cdot 32}

Subtract the numbers:

32 - 32 = 0

= \frac{0}{2 \cdot 32}

Multiply the numbers:

2 \cdot 32 = 64

= \frac{0}{64}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = 1, u = 0

Substitute back

u = sin (θ)

sin (θ) = 1, sin (θ) = 0

sin (θ) = 1, sin (θ) = 0

sin (θ) = 1 : θ = \frac{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{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 0

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 = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

4 - 4 sin (θ) = 4 cos (θ)

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

4 - 4 sin (θ) = 4 cos (θ)

plug in

θ = \frac{π}{2} + 2 π 1

4 - 4 sin (\frac{π}{2} + 2 π 1) = 4 cos (\frac{π}{2} + 2 π 1)

Refine

0 = 0

\Rightarrow True

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

4 - 4 sin (θ) = 4 cos (θ)

plug in

θ = 2 π 1

4 - 4 sin (2 π 1) = 4 cos (2 π 1)

Refine

4 = 4

\Rightarrow True

Check the solution

π + 2 πn :

False

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

4 - 4 sin (θ) = 4 cos (θ)

plug in

θ = π + 2 π 1

4 - 4 sin (π + 2 π 1) = 4 cos (π + 2 π 1)

Refine

4 = - 4

\Rightarrow False

θ = \frac{π}{2} + 2 πn, θ = 2 πn

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Frequently Asked Questions (FAQ)

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