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Popular >

sin^2(θ)(1+csc^2(θ))=k

Solution

sin^{2} (θ) (1 + csc^{2} (θ)) = k

Solution

θ = arcsin (k - 1) + 2 πn, θ = π + arcsin (- k - 1) + 2 πn, θ = arcsin (- k - 1) + 2 πn, θ = π + arcsin (k - 1) + 2 πn

Solution steps

sin^{2} (θ) (1 + csc^{2} (θ)) = k

Subtract

k

from both sides

sin^{2} (θ) (1 + csc^{2} (θ)) - k = 0

Rewrite using trig identities

- k + (1 + csc^{2} (θ)) sin^{2} (θ)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Simplify

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

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

(\frac{1}{sin ( θ )})^{2} = \frac{1}{sin ^{2} ( θ )}

(\frac{1}{sin ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sin ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

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

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

Expand

sin^{2} (θ) (1 + \frac{1}{sin ^{2} ( θ )}) : sin^{2} (θ) + 1

sin^{2} (θ) (1 + \frac{1}{sin ^{2} ( θ )})

Apply the distributive law:

a (b + c) = ab + a c

a = sin^{2} (θ), b = 1, c = \frac{1}{sin ^{2} ( θ )}

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

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

Simplify

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

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

1 \cdot sin^{2} (θ) = sin^{2} (θ)

1 \cdot sin^{2} (θ)

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

= sin^{2} (θ)

\frac{1}{sin ^{2} ( θ )} sin^{2} (θ) = 1

\frac{1}{sin ^{2} ( θ )} sin^{2} (θ)

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (θ)

= 1

= sin^{2} (θ) + 1

= sin^{2} (θ) + 1

= - k + sin^{2} (θ) + 1

= - k + sin^{2} (θ) + 1

1 + sin^{2} (θ) - k = 0

Solve by substitution

1 + sin^{2} (θ) - k = 0

Let:

sin (θ) = u

1 + u^{2} - k = 0

1 + u^{2} - k = 0 : u = k - 1, u = - k - 1

1 + u^{2} - k = 0

Move

k

to the right side

1 + u^{2} - k = 0

Add

k

to both sides

1 + u^{2} - k + k = 0 + k

Simplify

1 + u^{2} = k

1 + u^{2} = k

Move

1

to the right side

1 + u^{2} = k

Subtract

1

from both sides

1 + u^{2} - 1 = k - 1

Simplify

u^{2} = k - 1

u^{2} = k - 1

For

x^{2} = f (a)

the solutions are

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

u = k - 1, u = - k - 1

Substitute back

u = sin (θ)

sin (θ) = k - 1, sin (θ) = - k - 1

sin (θ) = k - 1, sin (θ) = - k - 1

sin (θ) = k - 1 : θ = arcsin (k - 1) + 2 πn, θ = π + arcsin (- k - 1) + 2 πn

sin (θ) = k - 1

Apply trig inverse properties

sin (θ) = k - 1

General solutions for

sin (θ) = k - 1

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

θ = arcsin (k - 1) + 2 πn, θ = π + arcsin (- k - 1) + 2 πn

θ = arcsin (k - 1) + 2 πn, θ = π + arcsin (- k - 1) + 2 πn

sin (θ) = - k - 1 : θ = arcsin (- k - 1) + 2 πn, θ = π + arcsin (k - 1) + 2 πn

sin (θ) = - k - 1

Apply trig inverse properties

sin (θ) = - k - 1

General solutions for

sin (θ) = - k - 1

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

θ = arcsin (- k - 1) + 2 πn, θ = π + arcsin (k - 1) + 2 πn

θ = arcsin (- k - 1) + 2 πn, θ = π + arcsin (k - 1) + 2 πn

Combine all the solutions

θ = arcsin (k - 1) + 2 πn, θ = π + arcsin (- k - 1) + 2 πn, θ = arcsin (- k - 1) + 2 πn, θ = π + arcsin (k - 1) + 2 πn

Graph

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