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

The general solution for 5cos^2(θ)=6sin(θ) is θ=0.60187…+2pin,θ=pi-0.60187…+2pin

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5cos^2(θ)=6sin(θ)

Solution

5 cos^{2} (θ) = 6 sin (θ)

Solution

θ = 0.60187 \dots + 2 πn, θ = π - 0.60187 \dots + 2 πn

Degrees

θ = 34.48499 \dots^{\circ} + 36 0^{\circ} n, θ = 145.51500 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 cos^{2} (θ) = 6 sin (θ)

Subtract

6 sin (θ)

from both sides

5 cos^{2} (θ) - 6 sin (θ) = 0

Rewrite using trig identities

5 cos^{2} (θ) - 6 sin (θ)

Use the Pythagorean identity:

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

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

= 5 (1 - sin^{2} (θ)) - 6 sin (θ)

(1 - sin^{2} (θ)) \cdot 5 - 6 sin (θ) = 0

Solve by substitution

(1 - sin^{2} (θ)) \cdot 5 - 6 sin (θ) = 0

Let:

sin (θ) = u

(1 - u^{2}) \cdot 5 - 6 u = 0

(1 - u^{2}) \cdot 5 - 6 u = 0 : u = - \frac{3 + 34}{5}, u = \frac{34 - 3}{5}

(1 - u^{2}) \cdot 5 - 6 u = 0

Expand

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

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

= 5 (1 - u^{2}) - 6 u

Expand

5 (1 - u^{2}) : 5 - 5 u^{2}

5 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 5, b = 1, c = u^{2}

= 5 \cdot 1 - 5 u^{2}

Multiply the numbers:

5 \cdot 1 = 5

= 5 - 5 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 6)^{2} - 4 (- 5) \cdot 5 = 234

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 6^{2} + 4 \cdot 5 \cdot 5

Multiply the numbers:

4 \cdot 5 \cdot 5 = 100

= 6^{2} + 100

6^{2} = 36

= 36 + 100

Add the numbers:

36 + 100 = 136

= 136

Prime factorization of

136 : 2^{3} \cdot 17

136

136

divides by

2136 = 68 \cdot 2

= 2 \cdot 68

68

divides by

268 = 34 \cdot 2

= 2 \cdot 2 \cdot 34

34

divides by

234 = 17 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 17

2, 17

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 17

= 2^{3} \cdot 17

= 2^{3} \cdot 17

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2 \cdot 17

Apply radical rule:

= 2^{2} 2 \cdot 17

Apply radical rule:

2^{2} = 2

= 2 2 \cdot 17

Refine

= 234

u_{1, 2} = \frac{- ( - 6 ) \pm 2 34}{2 ( - 5 )}

Separate the solutions

u_{1} = \frac{- ( - 6 ) + 2 34}{2 ( - 5 )}, u_{2} = \frac{- ( - 6 ) - 2 34}{2 ( - 5 )}

u = \frac{- ( - 6 ) + 2 34}{2 ( - 5 )} : - \frac{3 + 34}{5}

\frac{- ( - 6 ) + 2 34}{2 ( - 5 )}

Remove parentheses:

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

= \frac{6 + 2 34}{- 2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{6 + 2 34}{- 10}

Apply the fraction rule:

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

= - \frac{6 + 2 34}{10}

Cancel

\frac{6 + 2 34}{10} : \frac{3 + 34}{5}

\frac{6 + 2 34}{10}

Factor

6 + 234 : 2 (3 + 34)

6 + 234

Rewrite as

= 2 \cdot 3 + 234

Factor out common term

2

= 2 (3 + 34)

= \frac{2 ( 3 + 34 )}{10}

Cancel the common factor:

2

= \frac{3 + 34}{5}

= - \frac{3 + 34}{5}

u = \frac{- ( - 6 ) - 2 34}{2 ( - 5 )} : \frac{34 - 3}{5}

\frac{- ( - 6 ) - 2 34}{2 ( - 5 )}

Remove parentheses:

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

= \frac{6 - 2 34}{- 2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{6 - 2 34}{- 10}

Apply the fraction rule:

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

6 - 234 = - (234 - 6)

= \frac{2 34 - 6}{10}

Factor

234 - 6 : 2 (34 - 3)

234 - 6

Rewrite as

= 234 - 2 \cdot 3

Factor out common term

2

= 2 (34 - 3)

= \frac{2 ( 34 - 3 )}{10}

Cancel the common factor:

2

= \frac{34 - 3}{5}

The solutions to the quadratic equation are:

u = - \frac{3 + 34}{5}, u = \frac{34 - 3}{5}

Substitute back

u = sin (θ)

sin (θ) = - \frac{3 + 34}{5}, sin (θ) = \frac{34 - 3}{5}

sin (θ) = - \frac{3 + 34}{5}, sin (θ) = \frac{34 - 3}{5}

sin (θ) = - \frac{3 + 34}{5} :

No Solution

sin (θ) = - \frac{3 + 34}{5}

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = \frac{34 - 3}{5} : θ = arcsin (\frac{34 - 3}{5}) + 2 πn, θ = π - arcsin (\frac{34 - 3}{5}) + 2 πn

sin (θ) = \frac{34 - 3}{5}

Apply trig inverse properties

sin (θ) = \frac{34 - 3}{5}

General solutions for

sin (θ) = \frac{34 - 3}{5}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (\frac{34 - 3}{5}) + 2 πn, θ = π - arcsin (\frac{34 - 3}{5}) + 2 πn

θ = arcsin (\frac{34 - 3}{5}) + 2 πn, θ = π - arcsin (\frac{34 - 3}{5}) + 2 πn

Combine all the solutions

θ = arcsin (\frac{34 - 3}{5}) + 2 πn, θ = π - arcsin (\frac{34 - 3}{5}) + 2 πn

Show solutions in decimal form

θ = 0.60187 \dots + 2 πn, θ = π - 0.60187 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5cos^2(θ)=6sin(θ) ?
The general solution for 5cos^2(θ)=6sin(θ) is θ=0.60187…+2pin,θ=pi-0.60187…+2pin