What is the general solution for sin(-θ)=-3/5 ,tan(θ)>0 ?

The general solution for sin(-θ)=-3/5 ,tan(θ)>0 is No Solution for θ\in\mathbb{R}

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Popular >

sin(-θ)=-3/5 ,tan(θ)>0

Solution

sin (- θ) = - \frac{3}{5}, tan (θ) > 0

Solution

No Solution for θ \in R

Solution steps

sin (- θ) = - \frac{3}{5}, tan (θ) > 0

Apply trig inverse properties

sin (- θ) = - \frac{3}{5}

General solutions for

sin (- θ) = - \frac{3}{5}

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

- θ = arcsin (- \frac{3}{5}) + 2 πn, - θ = π + arcsin (\frac{3}{5}) + 2 πn

- θ = arcsin (- \frac{3}{5}) + 2 πn, - θ = π + arcsin (\frac{3}{5}) + 2 πn

Solve

- θ = arcsin (- \frac{3}{5}) + 2 πn : θ = arcsin (\frac{3}{5}) - 2 πn

- θ = arcsin (- \frac{3}{5}) + 2 πn

Simplify

arcsin (- \frac{3}{5}) + 2 πn : - arcsin (\frac{3}{5}) + 2 πn

arcsin (- \frac{3}{5}) + 2 πn

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{3}{5}) = - arcsin (\frac{3}{5})

= - arcsin (\frac{3}{5}) + 2 πn

- θ = - arcsin (\frac{3}{5}) + 2 πn

Divide both sides by

- 1

- θ = - arcsin (\frac{3}{5}) + 2 πn

Divide both sides by

- 1

\frac{- θ}{- 1} = - \frac{arcsin ( \frac{3}{5} )}{- 1} + \frac{2 πn}{- 1}

Simplify

\frac{- θ}{- 1} = - \frac{arcsin ( \frac{3}{5} )}{- 1} + \frac{2 πn}{- 1}

Simplify

\frac{- θ}{- 1} : θ

\frac{- θ}{- 1}

Apply the fraction rule:

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

= \frac{θ}{1}

Apply rule

\frac{a}{1} = a

= θ

Simplify

- \frac{arcsin ( \frac{3}{5} )}{- 1} + \frac{2 πn}{- 1} : arcsin (\frac{3}{5}) - 2 πn

- \frac{arcsin ( \frac{3}{5} )}{- 1} + \frac{2 πn}{- 1}

\frac{arcsin ( \frac{3}{5} )}{- 1} = - arcsin (\frac{3}{5})

\frac{arcsin ( \frac{3}{5} )}{- 1}

Apply the fraction rule:

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

= - \frac{arcsin ( \frac{3}{5} )}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{arcsin ( \frac{3}{5} )}{1} = arcsin (\frac{3}{5})

= - arcsin (\frac{3}{5})

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

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

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - (- arcsin (\frac{3}{5})) - 2 πn

Apply rule

- (- a) = a

= arcsin (\frac{3}{5}) - 2 πn

θ = arcsin (\frac{3}{5}) - 2 πn

θ = arcsin (\frac{3}{5}) - 2 πn

θ = arcsin (\frac{3}{5}) - 2 πn

Solve

- θ = π + arcsin (\frac{3}{5}) + 2 πn : θ = - π - arcsin (\frac{3}{5}) - 2 πn

- θ = π + arcsin (\frac{3}{5}) + 2 πn

Divide both sides by

- 1

- θ = π + arcsin (\frac{3}{5}) + 2 πn

Divide both sides by

- 1

\frac{- θ}{- 1} = \frac{π}{- 1} + \frac{arcsin ( \frac{3}{5} )}{- 1} + \frac{2 πn}{- 1}

Simplify

\frac{- θ}{- 1} = \frac{π}{- 1} + \frac{arcsin ( \frac{3}{5} )}{- 1} + \frac{2 πn}{- 1}

Simplify

\frac{- θ}{- 1} : θ

\frac{- θ}{- 1}

Apply the fraction rule:

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

= \frac{θ}{1}

Apply rule

\frac{a}{1} = a

= θ

Simplify

\frac{π}{- 1} + \frac{arcsin ( \frac{3}{5} )}{- 1} + \frac{2 πn}{- 1} : - π - arcsin (\frac{3}{5}) - 2 πn

\frac{π}{- 1} + \frac{arcsin ( \frac{3}{5} )}{- 1} + \frac{2 πn}{- 1}

\frac{π}{- 1} = - π

\frac{π}{- 1}

Apply the fraction rule:

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

= - \frac{π}{1}

Apply rule

\frac{a}{1} = a

= - π

= - π + \frac{arcsin ( \frac{3}{5} )}{- 1} + \frac{2 πn}{- 1}

\frac{arcsin ( \frac{3}{5} )}{- 1} = - arcsin (\frac{3}{5})

\frac{arcsin ( \frac{3}{5} )}{- 1}

Apply the fraction rule:

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

= - \frac{arcsin ( \frac{3}{5} )}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{arcsin ( \frac{3}{5} )}{1} = arcsin (\frac{3}{5})

= - arcsin (\frac{3}{5})

= - π - arcsin (\frac{3}{5}) + \frac{2 πn}{- 1}

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

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

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - π - arcsin (\frac{3}{5}) - 2 πn

θ = - π - arcsin (\frac{3}{5}) - 2 πn

θ = - π - arcsin (\frac{3}{5}) - 2 πn

θ = - π - arcsin (\frac{3}{5}) - 2 πn

θ = arcsin (\frac{3}{5}) - 2 πn, θ = - π - arcsin (\frac{3}{5}) - 2 πn

Solutions for the range

tan (θ) > 0

No Solution for θ \in R

Popular Examples

tan^3(x)+tan^2(x)-3tan(x)-3=0

tan^{3} (x) + tan^{2} (x) - 3 tan (x) - 3 = 0

tan^2(x)+5=3sec^2(x)

tan^{2} (x) + 5 = 3 sec^{2} (x)

tan(x)= 7/2

tan (x) = \frac{7}{2}

6cos^2(2x)+10cos(4x)=12.05718

6 cos^{2} (2 x) + 10 cos (4 x) = 12.05718

cos(θ/2)=sqrt((1+3/5)/2)

cos (\frac{θ}{2}) = \frac{1 + \frac{3}{5}}{2}

Frequently Asked Questions (FAQ)

What is the general solution for sin(-θ)=-3/5 ,tan(θ)>0 ?
The general solution for sin(-θ)=-3/5 ,tan(θ)>0 is No Solution for θ\in\mathbb{R}