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

The general solution for 6-sin(θ)=cos(2θ) is No Solution for θ\in\mathbb{R}

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Popular Trigonometry >

6-sin(θ)=cos(2θ)

Solution

6 - sin (θ) = cos (2 θ)

Solution

No Solution for θ \in R

Solution steps

6 - sin (θ) = cos (2 θ)

Subtract

cos (2 θ)

from both sides

6 - sin (θ) - cos (2 θ) = 0

Rewrite using trig identities

6 - cos (2 θ) - sin (θ)

Use the Double Angle identity:

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

= 6 - (1 - 2 sin^{2} (θ)) - sin (θ)

Simplify

6 - (1 - 2 sin^{2} (θ)) - sin (θ) : 2 sin^{2} (θ) - sin (θ) + 5

6 - (1 - 2 sin^{2} (θ)) - sin (θ)

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

- (1 - 2 sin^{2} (θ))

Distribute parentheses

= - (1) - (- 2 sin^{2} (θ))

Apply minus-plus rules

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

= - 1 + 2 sin^{2} (θ)

= 6 - 1 + 2 sin^{2} (θ) - sin (θ)

Subtract the numbers:

6 - 1 = 5

= 2 sin^{2} (θ) - sin (θ) + 5

= 2 sin^{2} (θ) - sin (θ) + 5

5 - sin (θ) + 2 sin^{2} (θ) = 0

Solve by substitution

5 - sin (θ) + 2 sin^{2} (θ) = 0

Let:

sin (θ) = u

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

5 - u + 2 u^{2} = 0 : u = \frac{1}{4} + i \frac{39}{4}, u = \frac{1}{4} - i \frac{39}{4}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 1, c = 5

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

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

Simplify

(- 1)^{2} - 4 \cdot 2 \cdot 5 : 39 i

(- 1)^{2} - 4 \cdot 2 \cdot 5

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 5 = 40

4 \cdot 2 \cdot 5

Multiply the numbers:

4 \cdot 2 \cdot 5 = 40

= 40

= 1 - 40

Subtract the numbers:

1 - 40 = - 39

= - 39

Apply radical rule:

- a = - 1 a

- 39 = - 1 39

= - 1 39

Apply imaginary number rule:

- 1 = i

= 39 i

u_{1, 2} = \frac{- ( - 1 ) \pm 39 i}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 39 i}{2 \cdot 2}, u_{2} = \frac{- ( - 1 ) - 39 i}{2 \cdot 2}

u = \frac{- ( - 1 ) + 39 i}{2 \cdot 2} : \frac{1}{4} + i \frac{39}{4}

\frac{- ( - 1 ) + 39 i}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 39 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 + 39 i}{4}

Rewrite

\frac{1 + 39 i}{4}

in standard complex form:

\frac{1}{4} + \frac{39}{4} i

\frac{1 + 39 i}{4}

Apply the fraction rule:

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

\frac{1 + 39 i}{4} = \frac{1}{4} + \frac{39 i}{4}

= \frac{1}{4} + \frac{39 i}{4}

= \frac{1}{4} + \frac{39}{4} i

u = \frac{- ( - 1 ) - 39 i}{2 \cdot 2} : \frac{1}{4} - i \frac{39}{4}

\frac{- ( - 1 ) - 39 i}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 - 39 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 - 39 i}{4}

Rewrite

\frac{1 - 39 i}{4}

in standard complex form:

\frac{1}{4} - \frac{39}{4} i

\frac{1 - 39 i}{4}

Apply the fraction rule:

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

\frac{1 - 39 i}{4} = \frac{1}{4} - \frac{39 i}{4}

= \frac{1}{4} - \frac{39 i}{4}

= \frac{1}{4} - \frac{39}{4} i

The solutions to the quadratic equation are:

u = \frac{1}{4} + i \frac{39}{4}, u = \frac{1}{4} - i \frac{39}{4}

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{4} + i \frac{39}{4}, sin (θ) = \frac{1}{4} - i \frac{39}{4}

sin (θ) = \frac{1}{4} + i \frac{39}{4}, sin (θ) = \frac{1}{4} - i \frac{39}{4}

sin (θ) = \frac{1}{4} + i \frac{39}{4} :

No Solution

sin (θ) = \frac{1}{4} + i \frac{39}{4}

No Solution

sin (θ) = \frac{1}{4} - i \frac{39}{4} :

No Solution

sin (θ) = \frac{1}{4} - i \frac{39}{4}

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 6-sin(θ)=cos(2θ) ?
The general solution for 6-sin(θ)=cos(2θ) is No Solution for θ\in\mathbb{R}