What is the general solution for 5+cos^2(θ)-6cos(θ)=0 ?

The general solution for 5+cos^2(θ)-6cos(θ)=0 is θ=2pin

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Popular >

5+cos^2(θ)-6cos(θ)=0

Solution

5 + cos^{2} (θ) - 6 cos (θ) = 0

Solution

θ = 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

5 + cos^{2} (θ) - 6 cos (θ) = 0

Solve by substitution

5 + cos^{2} (θ) - 6 cos (θ) = 0

Let:

cos (θ) = u

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

5 + u^{2} - 6 u = 0 : u = 5, u = 1

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

Write in the standard form

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

u^{2} - 6 u + 5 = 0

Solve with the quadratic formula

u^{2} - 6 u + 5 = 0

Quadratic Equation Formula:

For

a = 1, b = - 6, c = 5

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 5 = 4

(- 6)^{2} - 4 \cdot 1 \cdot 5

Apply exponent rule:

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

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 1 \cdot 5

Multiply the numbers:

4 \cdot 1 \cdot 5 = 20

= 6^{2} - 20

6^{2} = 36

= 36 - 20

Subtract the numbers:

36 - 20 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

u_{1} = \frac{- ( - 6 ) + 4}{2 \cdot 1}, u_{2} = \frac{- ( - 6 ) - 4}{2 \cdot 1}

u = \frac{- ( - 6 ) + 4}{2 \cdot 1} : 5

\frac{- ( - 6 ) + 4}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 + 4}{2 \cdot 1}

Add the numbers:

6 + 4 = 10

= \frac{10}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{10}{2}

Divide the numbers:

\frac{10}{2} = 5

= 5

u = \frac{- ( - 6 ) - 4}{2 \cdot 1} : 1

\frac{- ( - 6 ) - 4}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 - 4}{2 \cdot 1}

Subtract the numbers:

6 - 4 = 2

= \frac{2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 5, u = 1

Substitute back

u = cos (θ)

cos (θ) = 5, cos (θ) = 1

cos (θ) = 5, cos (θ) = 1

cos (θ) = 5 :

No Solution

cos (θ) = 5

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

General solutions for

cos (θ) = 1

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 = 0 + 2 πn

θ = 0 + 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

Combine all the solutions

θ = 2 πn

Popular Examples

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tan (2 c) = - tan (x)

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

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

sin(θ)-0.1cos(θ)=(8.87)/(9.8)

sin (θ) - 0.1 cos (θ) = \frac{8.87}{9.8}

2sin(x)=(sin(x))/(cos(x))

2 sin (x) = \frac{sin ( x )}{cos ( x )}

6sin^2(x)=5+cos(x),180<= x<= 540

6 sin^{2} (x) = 5 + cos (x), 18 0^{\circ} \leq x \leq 54 0^{\circ}

Frequently Asked Questions (FAQ)

What is the general solution for 5+cos^2(θ)-6cos(θ)=0 ?
The general solution for 5+cos^2(θ)-6cos(θ)=0 is θ=2pin