What is the general solution for 7cos^2(θ)+6sin(θ)-10=-4 ?

The general solution for 7cos^2(θ)+6sin(θ)-10=-4 is θ=-0.14334…+2pin,θ=pi+0.14334…+2pin,θ= pi/2+2pin

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7cos^2(θ)+6sin(θ)-10=-4

Solution

7 cos^{2} (θ) + 6 sin (θ) - 10 = - 4

Solution

θ = - 0.14334 \dots + 2 πn, θ = π + 0.14334 \dots + 2 πn, θ = \frac{π}{2} + 2 πn

Degrees

θ = - 8.21321 \dots^{\circ} + 36 0^{\circ} n, θ = 188.21321 \dots^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

7 cos^{2} (θ) + 6 sin (θ) - 10 = - 4

Subtract

- 4

from both sides

7 cos^{2} (θ) + 6 sin (θ) - 6 = 0

Rewrite using trig identities

- 6 + 6 sin (θ) + 7 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= - 6 + 6 sin (θ) + 7 (1 - sin^{2} (θ))

Simplify

- 6 + 6 sin (θ) + 7 (1 - sin^{2} (θ)) : 6 sin (θ) - 7 sin^{2} (θ) + 1

- 6 + 6 sin (θ) + 7 (1 - sin^{2} (θ))

Expand

7 (1 - sin^{2} (θ)) : 7 - 7 sin^{2} (θ)

7 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 7, b = 1, c = sin^{2} (θ)

= 7 \cdot 1 - 7 sin^{2} (θ)

Multiply the numbers:

7 \cdot 1 = 7

= 7 - 7 sin^{2} (θ)

= - 6 + 6 sin (θ) + 7 - 7 sin^{2} (θ)

Simplify

- 6 + 6 sin (θ) + 7 - 7 sin^{2} (θ) : 6 sin (θ) - 7 sin^{2} (θ) + 1

- 6 + 6 sin (θ) + 7 - 7 sin^{2} (θ)

Group like terms

= 6 sin (θ) - 7 sin^{2} (θ) - 6 + 7

Add/Subtract the numbers:

- 6 + 7 = 1

= 6 sin (θ) - 7 sin^{2} (θ) + 1

= 6 sin (θ) - 7 sin^{2} (θ) + 1

= 6 sin (θ) - 7 sin^{2} (θ) + 1

1 + 6 sin (θ) - 7 sin^{2} (θ) = 0

Solve by substitution

1 + 6 sin (θ) - 7 sin^{2} (θ) = 0

Let:

sin (θ) = u

1 + 6 u - 7 u^{2} = 0

1 + 6 u - 7 u^{2} = 0 : u = - \frac{1}{7}, u = 1

1 + 6 u - 7 u^{2} = 0

Write in the standard form

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

- 7 u^{2} + 6 u + 1 = 0

Solve with the quadratic formula

- 7 u^{2} + 6 u + 1 = 0

Quadratic Equation Formula:

For

a = - 7, b = 6, c = 1

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 7 ) \cdot 1}{2 ( - 7 )}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 7 ) \cdot 1}{2 ( - 7 )}

6^{2} - 4 (- 7) \cdot 1 = 8

6^{2} - 4 (- 7) \cdot 1

Apply rule

- (- a) = a

= 6^{2} + 4 \cdot 7 \cdot 1

Multiply the numbers:

4 \cdot 7 \cdot 1 = 28

= 6^{2} + 28

6^{2} = 36

= 36 + 28

Add the numbers:

36 + 28 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- 6 \pm 8}{2 ( - 7 )}

Separate the solutions

u_{1} = \frac{- 6 + 8}{2 ( - 7 )}, u_{2} = \frac{- 6 - 8}{2 ( - 7 )}

u = \frac{- 6 + 8}{2 ( - 7 )} : - \frac{1}{7}

\frac{- 6 + 8}{2 ( - 7 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 + 8}{- 2 \cdot 7}

Add/Subtract the numbers:

- 6 + 8 = 2

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

Multiply the numbers:

2 \cdot 7 = 14

= \frac{2}{- 14}

Apply the fraction rule:

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

= - \frac{2}{14}

Cancel the common factor:

2

= - \frac{1}{7}

u = \frac{- 6 - 8}{2 ( - 7 )} : 1

\frac{- 6 - 8}{2 ( - 7 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 - 8}{- 2 \cdot 7}

Subtract the numbers:

- 6 - 8 = - 14

= \frac{- 14}{- 2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{- 14}{- 14}

Apply the fraction rule:

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

= \frac{14}{14}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - \frac{1}{7}, u = 1

Substitute back

u = sin (θ)

sin (θ) = - \frac{1}{7}, sin (θ) = 1

sin (θ) = - \frac{1}{7}, sin (θ) = 1

sin (θ) = - \frac{1}{7} : θ = arcsin (- \frac{1}{7}) + 2 πn, θ = π + arcsin (\frac{1}{7}) + 2 πn

sin (θ) = - \frac{1}{7}

Apply trig inverse properties

sin (θ) = - \frac{1}{7}

General solutions for

sin (θ) = - \frac{1}{7}

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

θ = arcsin (- \frac{1}{7}) + 2 πn, θ = π + arcsin (\frac{1}{7}) + 2 πn

θ = arcsin (- \frac{1}{7}) + 2 πn, θ = π + arcsin (\frac{1}{7}) + 2 πn

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = 1

General solutions for

sin (θ) = 1

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

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Combine all the solutions

θ = arcsin (- \frac{1}{7}) + 2 πn, θ = π + arcsin (\frac{1}{7}) + 2 πn, θ = \frac{π}{2} + 2 πn

Show solutions in decimal form

θ = - 0.14334 \dots + 2 πn, θ = π + 0.14334 \dots + 2 πn, θ = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 7cos^2(θ)+6sin(θ)-10=-4 ?
The general solution for 7cos^2(θ)+6sin(θ)-10=-4 is θ=-0.14334…+2pin,θ=pi+0.14334…+2pin,θ= pi/2+2pin