What is the general solution for 7cos(2θ)-9=-9sin(θ) ?

The general solution for 7cos(2θ)-9=-9sin(θ) is No Solution for θ\in\mathbb{R}

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Popular Trigonometry >

7cos(2θ)-9=-9sin(θ)

Solution

7 cos (2 θ) - 9 = - 9 sin (θ)

Solution

No Solution for θ \in R

Solution steps

7 cos (2 θ) - 9 = - 9 sin (θ)

Subtract

- 9 sin (θ)

from both sides

7 cos (2 θ) - 9 + 9 sin (θ) = 0

Rewrite using trig identities

- 9 + 7 cos (2 θ) + 9 sin (θ)

Use the Double Angle identity:

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

= - 9 + 7 (1 - 2 sin^{2} (θ)) + 9 sin (θ)

Simplify

- 9 + 7 (1 - 2 sin^{2} (θ)) + 9 sin (θ) : 9 sin (θ) - 14 sin^{2} (θ) - 2

- 9 + 7 (1 - 2 sin^{2} (θ)) + 9 sin (θ)

Expand

7 (1 - 2 sin^{2} (θ)) : 7 - 14 sin^{2} (θ)

7 (1 - 2 sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 7, b = 1, c = 2 sin^{2} (θ)

= 7 \cdot 1 - 7 \cdot 2 sin^{2} (θ)

Simplify

7 \cdot 1 - 7 \cdot 2 sin^{2} (θ) : 7 - 14 sin^{2} (θ)

7 \cdot 1 - 7 \cdot 2 sin^{2} (θ)

Multiply the numbers:

7 \cdot 1 = 7

= 7 - 7 \cdot 2 sin^{2} (θ)

Multiply the numbers:

7 \cdot 2 = 14

= 7 - 14 sin^{2} (θ)

= 7 - 14 sin^{2} (θ)

= - 9 + 7 - 14 sin^{2} (θ) + 9 sin (θ)

Add/Subtract the numbers:

- 9 + 7 = - 2

= 9 sin (θ) - 14 sin^{2} (θ) - 2

= 9 sin (θ) - 14 sin^{2} (θ) - 2

- 2 - 14 sin^{2} (θ) + 9 sin (θ) = 0

Solve by substitution

- 2 - 14 sin^{2} (θ) + 9 sin (θ) = 0

Let:

sin (θ) = u

- 2 - 14 u^{2} + 9 u = 0

- 2 - 14 u^{2} + 9 u = 0 : u = \frac{9}{28} - i \frac{31}{28}, u = \frac{9}{28} + i \frac{31}{28}

- 2 - 14 u^{2} + 9 u = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 14, b = 9, c = - 2

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 ( - 14 ) ( - 2 )}{2 ( - 14 )}

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 ( - 14 ) ( - 2 )}{2 ( - 14 )}

Simplify

9^{2} - 4 (- 14) (- 2) : 31 i

9^{2} - 4 (- 14) (- 2)

Apply rule

- (- a) = a

= 9^{2} - 4 \cdot 14 \cdot 2

Multiply the numbers:

4 \cdot 14 \cdot 2 = 112

= 9^{2} - 112

Apply imaginary number rule:

- a = i a

= i 112 - 9^{2}

- 9^{2} + 112 = 31

- 9^{2} + 112

9^{2} = 81

= - 81 + 112

Add/Subtract the numbers:

- 81 + 112 = 31

= 31

= 31 i

u_{1, 2} = \frac{- 9 \pm 31 i}{2 ( - 14 )}

Separate the solutions

u_{1} = \frac{- 9 + 31 i}{2 ( - 14 )}, u_{2} = \frac{- 9 - 31 i}{2 ( - 14 )}

u = \frac{- 9 + 31 i}{2 ( - 14 )} : \frac{9}{28} - i \frac{31}{28}

\frac{- 9 + 31 i}{2 ( - 14 )}

Remove parentheses:

(- a) = - a

= \frac{- 9 + 31 i}{- 2 \cdot 14}

Multiply the numbers:

2 \cdot 14 = 28

= \frac{- 9 + 31 i}{- 28}

Apply the fraction rule:

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

= - \frac{- 9 + 31 i}{28}

Rewrite

- \frac{- 9 + 31 i}{28}

in standard complex form:

\frac{9}{28} - \frac{31}{28} i

- \frac{- 9 + 31 i}{28}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 9 + 31 i}{28} = - (- \frac{9}{28}) - (\frac{31 i}{28})

= - (- \frac{9}{28}) - (\frac{31 i}{28})

Remove parentheses:

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

= \frac{9}{28} - \frac{31 i}{28}

= \frac{9}{28} - \frac{31}{28} i

u = \frac{- 9 - 31 i}{2 ( - 14 )} : \frac{9}{28} + i \frac{31}{28}

\frac{- 9 - 31 i}{2 ( - 14 )}

Remove parentheses:

(- a) = - a

= \frac{- 9 - 31 i}{- 2 \cdot 14}

Multiply the numbers:

2 \cdot 14 = 28

= \frac{- 9 - 31 i}{- 28}

Apply the fraction rule:

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

= - \frac{- 9 - 31 i}{28}

Rewrite

- \frac{- 9 - 31 i}{28}

in standard complex form:

\frac{9}{28} + \frac{31}{28} i

- \frac{- 9 - 31 i}{28}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 9 - 31 i}{28} = - (- \frac{9}{28}) - (- \frac{31 i}{28})

= - (- \frac{9}{28}) - (- \frac{31 i}{28})

Apply rule

- (- a) = a

= \frac{9}{28} + \frac{31 i}{28}

= \frac{9}{28} + \frac{31}{28} i

The solutions to the quadratic equation are:

u = \frac{9}{28} - i \frac{31}{28}, u = \frac{9}{28} + i \frac{31}{28}

Substitute back

u = sin (θ)

sin (θ) = \frac{9}{28} - i \frac{31}{28}, sin (θ) = \frac{9}{28} + i \frac{31}{28}

sin (θ) = \frac{9}{28} - i \frac{31}{28}, sin (θ) = \frac{9}{28} + i \frac{31}{28}

sin (θ) = \frac{9}{28} - i \frac{31}{28} :

No Solution

sin (θ) = \frac{9}{28} - i \frac{31}{28}

No Solution

sin (θ) = \frac{9}{28} + i \frac{31}{28} :

No Solution

sin (θ) = \frac{9}{28} + i \frac{31}{28}

No Solution

Combine all the solutions

No Solution for θ \in R

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 7cos(2θ)-9=-9sin(θ) ?
The general solution for 7cos(2θ)-9=-9sin(θ) is No Solution for θ\in\mathbb{R}