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Popular Trigonometry >

2cos^4(θ)+9sin^2(θ)=5

Solution

2 cos^{4} (θ) + 9 sin^{2} (θ) = 5

Solution

θ = 0.78539 \dots + 2 πn, θ = 2 π - 0.78539 \dots + 2 πn, θ = 2.35619 \dots + 2 πn, θ = - 2.35619 \dots + 2 πn

Degrees

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 36 0^{\circ} n, θ = - 13 5^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{4} (θ) + 9 sin^{2} (θ) = 5

Subtract

5

from both sides

2 cos^{4} (θ) + 9 sin^{2} (θ) - 5 = 0

Rewrite using trig identities

- 5 + 2 cos^{4} (θ) + 9 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= - 5 + 2 cos^{4} (θ) + 9 (1 - cos^{2} (θ))

Simplify

- 5 + 2 cos^{4} (θ) + 9 (1 - cos^{2} (θ)) : 2 cos^{4} (θ) - 9 cos^{2} (θ) + 4

- 5 + 2 cos^{4} (θ) + 9 (1 - cos^{2} (θ))

Expand

9 (1 - cos^{2} (θ)) : 9 - 9 cos^{2} (θ)

9 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 9, b = 1, c = cos^{2} (θ)

= 9 \cdot 1 - 9 cos^{2} (θ)

Multiply the numbers:

9 \cdot 1 = 9

= 9 - 9 cos^{2} (θ)

= - 5 + 2 cos^{4} (θ) + 9 - 9 cos^{2} (θ)

Simplify

- 5 + 2 cos^{4} (θ) + 9 - 9 cos^{2} (θ) : 2 cos^{4} (θ) - 9 cos^{2} (θ) + 4

- 5 + 2 cos^{4} (θ) + 9 - 9 cos^{2} (θ)

Group like terms

= 2 cos^{4} (θ) - 9 cos^{2} (θ) - 5 + 9

Add/Subtract the numbers:

- 5 + 9 = 4

= 2 cos^{4} (θ) - 9 cos^{2} (θ) + 4

= 2 cos^{4} (θ) - 9 cos^{2} (θ) + 4

= 2 cos^{4} (θ) - 9 cos^{2} (θ) + 4

4 + 2 cos^{4} (θ) - 9 cos^{2} (θ) = 0

Solve by substitution

4 + 2 cos^{4} (θ) - 9 cos^{2} (θ) = 0

Let:

cos (θ) = u

4 + 2 u^{4} - 9 u^{2} = 0

4 + 2 u^{4} - 9 u^{2} = 0 : u = 2, u = - 2, u = \frac{1}{2}, u = - \frac{1}{2}

4 + 2 u^{4} - 9 u^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

2 u^{4} - 9 u^{2} + 4 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

2 v^{2} - 9 v + 4 = 0

Solve

2 v^{2} - 9 v + 4 = 0 : v = 4, v = \frac{1}{2}

2 v^{2} - 9 v + 4 = 0

Solve with the quadratic formula

2 v^{2} - 9 v + 4 = 0

Quadratic Equation Formula:

For

a = 2, b = - 9, c = 4

v_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

v_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

(- 9)^{2} - 4 \cdot 2 \cdot 4 = 7

(- 9)^{2} - 4 \cdot 2 \cdot 4

Apply exponent rule:

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

n

is even

(- 9)^{2} = 9^{2}

= 9^{2} - 4 \cdot 2 \cdot 4

Multiply the numbers:

4 \cdot 2 \cdot 4 = 32

= 9^{2} - 32

9^{2} = 81

= 81 - 32

Subtract the numbers:

81 - 32 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

v_{1, 2} = \frac{- ( - 9 ) \pm 7}{2 \cdot 2}

Separate the solutions

v_{1} = \frac{- ( - 9 ) + 7}{2 \cdot 2}, v_{2} = \frac{- ( - 9 ) - 7}{2 \cdot 2}

v = \frac{- ( - 9 ) + 7}{2 \cdot 2} : 4

\frac{- ( - 9 ) + 7}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{9 + 7}{2 \cdot 2}

Add the numbers:

9 + 7 = 16

= \frac{16}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{16}{4}

Divide the numbers:

\frac{16}{4} = 4

= 4

v = \frac{- ( - 9 ) - 7}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 9 ) - 7}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{9 - 7}{2 \cdot 2}

Subtract the numbers:

9 - 7 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

v = 4, v = \frac{1}{2}

v = 4, v = \frac{1}{2}

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = 4 : u = 2, u = - 2

u^{2} = 4

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 4, u = - 4

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

- 4 = - 2

- 4

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= - 2

u = 2, u = - 2

Solve

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

u^{2} = \frac{1}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

The solutions are

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

Substitute back

u = cos (θ)

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

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

cos (θ) = 2 :

No Solution

cos (θ) = 2

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = - 2 :

No Solution

cos (θ) = - 2

- 1 \leq cos (x) \leq 1

No Solution

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

cos (θ) = \frac{1}{2}

Apply trig inverse properties

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn

θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (θ) = - \frac{1}{2} : θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

cos (θ) = - \frac{1}{2}

Apply trig inverse properties

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

Combine all the solutions

θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn, θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

Show solutions in decimal form

θ = 0.78539 \dots + 2 πn, θ = 2 π - 0.78539 \dots + 2 πn, θ = 2.35619 \dots + 2 πn, θ = - 2.35619 \dots + 2 πn

Graph

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