What is the general solution for sin(θ)-cos^2(θ)sin(θ)=-1 ?

The general solution for sin(θ)-cos^2(θ)sin(θ)=-1 is θ=(3pi)/2+2pin

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Popular >

sin(θ)-cos^2(θ)sin(θ)=-1

Solution

sin (θ) - cos^{2} (θ) sin (θ) = - 1

Solution

θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ) - cos^{2} (θ) sin (θ) = - 1

Subtract

- 1

from both sides

sin (θ) - cos^{2} (θ) sin (θ) + 1 = 0

Rewrite using trig identities

1 + sin (θ) - cos^{2} (θ) sin (θ)

Use the Pythagorean identity:

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

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

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

Simplify

1 + sin (θ) - (1 - sin^{2} (θ)) sin (θ) : 1 + sin^{3} (θ)

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

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

Expand

- sin (θ) (1 - sin^{2} (θ)) : - sin (θ) + sin^{3} (θ)

- sin (θ) (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - sin (θ), b = 1, c = sin^{2} (θ)

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

Apply minus-plus rules

- (- a) = a

= - 1 \cdot sin (θ) + sin^{2} (θ) sin (θ)

Simplify

- 1 \cdot sin (θ) + sin^{2} (θ) sin (θ) : - sin (θ) + sin^{3} (θ)

- 1 \cdot sin (θ) + sin^{2} (θ) sin (θ)

1 \cdot sin (θ) = sin (θ)

1 \cdot sin (θ)

Multiply:

1 \cdot sin (θ) = sin (θ)

= sin (θ)

sin^{2} (θ) sin (θ) = sin^{3} (θ)

sin^{2} (θ) sin (θ)

Apply exponent rule:

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

sin^{2} (θ) sin (θ) = sin^{2 + 1} (θ)

= sin^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= sin^{3} (θ)

= - sin (θ) + sin^{3} (θ)

= - sin (θ) + sin^{3} (θ)

= 1 + sin (θ) - sin (θ) + sin^{3} (θ)

Add similar elements:

sin (θ) - sin (θ) = 0

= 1 + sin^{3} (θ)

= 1 + sin^{3} (θ)

1 + sin^{3} (θ) = 0

Solve by substitution

1 + sin^{3} (θ) = 0

Let:

sin (θ) = u

1 + u^{3} = 0

1 + u^{3} = 0 : u = - 1, u = \frac{1}{2} - i \frac{3}{2}, u = \frac{1}{2} + i \frac{3}{2}

1 + u^{3} = 0

Move

1

to the right side

1 + u^{3} = 0

Subtract

1

from both sides

1 + u^{3} - 1 = 0 - 1

Simplify

u^{3} = - 1

u^{3} = - 1

For

x^{3} = f (a)

the solutions are

x = 3 f (a), 3 f (a) \frac{- 1 - 3 i}{2}, 3 f (a) \frac{- 1 + 3 i}{2}

u = 3 - 1, u = 3 - 1 \frac{- 1 + 3 i}{2}, u = 3 - 1 \frac{- 1 - 3 i}{2}

3 - 1 = - 1

3 - 1

n - 1 = - 1,

n

is odd

3 - 1 = - 1

= - 1

Simplify

3 - 1 \frac{- 1 + 3 i}{2} : \frac{1}{2} - i \frac{3}{2}

3 - 1 \frac{- 1 + 3 i}{2}

3 - 1 = - 1

3 - 1

n - 1 = - 1,

n

is odd

3 - 1 = - 1

= - 1

= - 1 \cdot \frac{- 1 + 3 i}{2}

Multiply:

1 \cdot \frac{- 1 + 3 i}{2} = \frac{- 1 + 3 i}{2}

= - \frac{- 1 + 3 i}{2}

Rewrite

- \frac{- 1 + 3 i}{2}

in standard complex form:

\frac{1}{2} - \frac{3}{2} i

- \frac{- 1 + 3 i}{2}

Apply the fraction rule:

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

\frac{- 1 + 3 i}{2} = - (- \frac{1}{2}) - (\frac{3 i}{2})

= - (- \frac{1}{2}) - (\frac{3 i}{2})

Remove parentheses:

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

= \frac{1}{2} - \frac{3 i}{2}

= \frac{1}{2} - \frac{3}{2} i

Simplify

3 - 1 \frac{- 1 - 3 i}{2} : \frac{1}{2} + i \frac{3}{2}

3 - 1 \frac{- 1 - 3 i}{2}

3 - 1 = - 1

3 - 1

n - 1 = - 1,

n

is odd

3 - 1 = - 1

= - 1

= - 1 \cdot \frac{- 1 - 3 i}{2}

Multiply:

1 \cdot \frac{- 1 - 3 i}{2} = \frac{- 1 - 3 i}{2}

= - \frac{- 1 - 3 i}{2}

Rewrite

- \frac{- 1 - 3 i}{2}

in standard complex form:

\frac{1}{2} + \frac{3}{2} i

- \frac{- 1 - 3 i}{2}

Apply the fraction rule:

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

\frac{- 1 - 3 i}{2} = - (- \frac{1}{2}) - (- \frac{3 i}{2})

= - (- \frac{1}{2}) - (- \frac{3 i}{2})

Apply rule

- (- a) = a

= \frac{1}{2} + \frac{3 i}{2}

= \frac{1}{2} + \frac{3}{2} i

u = - 1, u = \frac{1}{2} - i \frac{3}{2}, u = \frac{1}{2} + i \frac{3}{2}

Substitute back

u = sin (θ)

sin (θ) = - 1, sin (θ) = \frac{1}{2} - i \frac{3}{2}, sin (θ) = \frac{1}{2} + i \frac{3}{2}

sin (θ) = - 1, sin (θ) = \frac{1}{2} - i \frac{3}{2}, sin (θ) = \frac{1}{2} + i \frac{3}{2}

sin (θ) = - 1 : θ = \frac{3 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{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

sin (θ) = \frac{1}{2} - i \frac{3}{2} :

No Solution

sin (θ) = \frac{1}{2} - i \frac{3}{2}

No Solution

sin (θ) = \frac{1}{2} + i \frac{3}{2} :

No Solution

sin (θ) = \frac{1}{2} + i \frac{3}{2}

No Solution

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sin(θ)-cos^2(θ)sin(θ)=-1 ?
The general solution for sin(θ)-cos^2(θ)sin(θ)=-1 is θ=(3pi)/2+2pin