What is the general solution for 10cos(θ)=4sin^2(θ)+2 ?

The general solution for 10cos(θ)=4sin^2(θ)+2 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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10cos(θ)=4sin^2(θ)+2

Solution

10 cos (θ) = 4 sin^{2} (θ) + 2

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

10 cos (θ) = 4 sin^{2} (θ) + 2

Subtract

4 sin^{2} (θ) + 2

from both sides

10 cos (θ) - 4 sin^{2} (θ) - 2 = 0

Rewrite using trig identities

- 2 + 10 cos (θ) - 4 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= - 2 + 10 cos (θ) - 4 (1 - cos^{2} (θ))

Simplify

- 2 + 10 cos (θ) - 4 (1 - cos^{2} (θ)) : 4 cos^{2} (θ) + 10 cos (θ) - 6

- 2 + 10 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)

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 cos^{2} (θ)

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 cos^{2} (θ)

= - 2 + 10 cos (θ) - 4 + 4 cos^{2} (θ)

Simplify

- 2 + 10 cos (θ) - 4 + 4 cos^{2} (θ) : 4 cos^{2} (θ) + 10 cos (θ) - 6

- 2 + 10 cos (θ) - 4 + 4 cos^{2} (θ)

Group like terms

= 10 cos (θ) + 4 cos^{2} (θ) - 2 - 4

Subtract the numbers:

- 2 - 4 = - 6

= 4 cos^{2} (θ) + 10 cos (θ) - 6

= 4 cos^{2} (θ) + 10 cos (θ) - 6

= 4 cos^{2} (θ) + 10 cos (θ) - 6

- 6 + 10 cos (θ) + 4 cos^{2} (θ) = 0

Solve by substitution

- 6 + 10 cos (θ) + 4 cos^{2} (θ) = 0

Let:

cos (θ) = u

- 6 + 10 u + 4 u^{2} = 0

- 6 + 10 u + 4 u^{2} = 0 : u = \frac{1}{2}, u = - 3

- 6 + 10 u + 4 u^{2} = 0

Write in the standard form

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

4 u^{2} + 10 u - 6 = 0

Solve with the quadratic formula

4 u^{2} + 10 u - 6 = 0

Quadratic Equation Formula:

For

a = 4, b = 10, c = - 6

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 4 ( - 6 )}{2 \cdot 4}

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 4 ( - 6 )}{2 \cdot 4}

1 0^{2} - 4 \cdot 4 (- 6) = 14

1 0^{2} - 4 \cdot 4 (- 6)

Apply rule

- (- a) = a

= 1 0^{2} + 4 \cdot 4 \cdot 6

Multiply the numbers:

4 \cdot 4 \cdot 6 = 96

= 1 0^{2} + 96

1 0^{2} = 100

= 100 + 96

Add the numbers:

100 + 96 = 196

= 196

Factor the number:

196 = 1 4^{2}

= 1 4^{2}

Apply radical rule:

1 4^{2} = 14

= 14

u_{1, 2} = \frac{- 10 \pm 14}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 10 + 14}{2 \cdot 4}, u_{2} = \frac{- 10 - 14}{2 \cdot 4}

u = \frac{- 10 + 14}{2 \cdot 4} : \frac{1}{2}

\frac{- 10 + 14}{2 \cdot 4}

Add/Subtract the numbers:

- 10 + 14 = 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{- 10 - 14}{2 \cdot 4} : - 3

\frac{- 10 - 14}{2 \cdot 4}

Subtract the numbers:

- 10 - 14 = - 24

= \frac{- 24}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 24}{8}

Apply the fraction rule:

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

= - \frac{24}{8}

Divide the numbers:

\frac{24}{8} = 3

= - 3

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = \frac{1}{2}, cos (θ) = - 3

cos (θ) = \frac{1}{2}, cos (θ) = - 3

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

cos (θ) = - 3 :

No Solution

cos (θ) = - 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

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csc(x)= 1/(cos(x))

csc (x) = \frac{1}{cos ( x )}

csc(θ)=(1.201)

csc (θ) = (1.201)

Frequently Asked Questions (FAQ)

What is the general solution for 10cos(θ)=4sin^2(θ)+2 ?
The general solution for 10cos(θ)=4sin^2(θ)+2 is θ= pi/3+2pin,θ=(5pi)/3+2pin