What is the general solution for 4sin^2(θ)-4cos(θ)=1 ?

The general solution for 4sin^2(θ)-4cos(θ)=1 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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Popular >

4sin^2(θ)-4cos(θ)=1

Solution

4 sin^{2} (θ) - 4 cos (θ) = 1

Solution

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

Degrees

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

4 sin^{2} (θ) - 4 cos (θ) = 1

Subtract

1

from both sides

4 sin^{2} (θ) - 4 cos (θ) - 1 = 0

Rewrite using trig identities

- 1 - 4 cos (θ) + 4 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= - 1 - 4 cos (θ) + 4 (1 - cos^{2} (θ))

Simplify

- 1 - 4 cos (θ) + 4 (1 - cos^{2} (θ)) : - 4 cos^{2} (θ) - 4 cos (θ) + 3

- 1 - 4 cos (θ) + 4 (1 - cos^{2} (θ))

Expand

4 (1 - cos^{2} (θ)) : 4 - 4 cos^{2} (θ)

4 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (θ)

= 4 \cdot 1 - 4 cos^{2} (θ)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (θ)

= - 1 - 4 cos (θ) + 4 - 4 cos^{2} (θ)

Simplify

- 1 - 4 cos (θ) + 4 - 4 cos^{2} (θ) : - 4 cos^{2} (θ) - 4 cos (θ) + 3

- 1 - 4 cos (θ) + 4 - 4 cos^{2} (θ)

Group like terms

= - 4 cos (θ) - 4 cos^{2} (θ) - 1 + 4

Add/Subtract the numbers:

- 1 + 4 = 3

= - 4 cos^{2} (θ) - 4 cos (θ) + 3

= - 4 cos^{2} (θ) - 4 cos (θ) + 3

= - 4 cos^{2} (θ) - 4 cos (θ) + 3

3 - 4 cos (θ) - 4 cos^{2} (θ) = 0

Solve by substitution

3 - 4 cos (θ) - 4 cos^{2} (θ) = 0

Let:

cos (θ) = u

3 - 4 u - 4 u^{2} = 0

3 - 4 u - 4 u^{2} = 0 : u = - \frac{3}{2}, u = \frac{1}{2}

3 - 4 u - 4 u^{2} = 0

Write in the standard form

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

- 4 u^{2} - 4 u + 3 = 0

Solve with the quadratic formula

- 4 u^{2} - 4 u + 3 = 0

Quadratic Equation Formula:

For

a = - 4, b = - 4, c = 3

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

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

(- 4)^{2} - 4 (- 4) \cdot 3 = 8

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2}

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

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 4^{2} + 48

4^{2} = 16

= 16 + 48

Add the numbers:

16 + 48 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

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

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 8}{2 ( - 4 )}, u_{2} = \frac{- ( - 4 ) - 8}{2 ( - 4 )}

u = \frac{- ( - 4 ) + 8}{2 ( - 4 )} : - \frac{3}{2}

\frac{- ( - 4 ) + 8}{2 ( - 4 )}

Remove parentheses:

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

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

Add the numbers:

4 + 8 = 12

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{12}{- 8}

Apply the fraction rule:

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

= - \frac{12}{8}

Cancel the common factor:

4

= - \frac{3}{2}

u = \frac{- ( - 4 ) - 8}{2 ( - 4 )} : \frac{1}{2}

\frac{- ( - 4 ) - 8}{2 ( - 4 )}

Remove parentheses:

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

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

Subtract the numbers:

4 - 8 = - 4

= \frac{- 4}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

Apply the fraction rule:

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

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

The solutions to the quadratic equation are:

u = - \frac{3}{2}, u = \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = - \frac{3}{2}, cos (θ) = \frac{1}{2}

cos (θ) = - \frac{3}{2}, cos (θ) = \frac{1}{2}

cos (θ) = - \frac{3}{2} :

No Solution

cos (θ) = - \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

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

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

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

Combine all the solutions

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

Popular Examples

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\frac{33}{sin ( x )} = \frac{29}{sin ( 10 0 ^{\circ} )}

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

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

(1/2)=cos(θ)

(\frac{1}{2}) = cos (θ)

sin(θ)= 4/5 ,θ,I,cos(2θ)

sin (θ) = \frac{4}{5}, θ, I, cos (2 θ)

sec^2(x)cot(x)=4

sec^{2} (x) cot (x) = 4

Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(θ)-4cos(θ)=1 ?
The general solution for 4sin^2(θ)-4cos(θ)=1 is θ= pi/3+2pin,θ=(5pi)/3+2pin