What is the general solution for sin(b)=((250km)/(590km))sin(23),82 ?

The general solution for sin(b)=((250km)/(590km))sin(23),82 is No Solution for b\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(b)=((250km)/(590km))sin(23),82

sin (b) = (\frac{250 km}{590 km}) sin (2 3^{\circ}), 8 2^{\circ}

No Solution for b \in R

Solution steps

sin (b) = (\frac{250 km}{590 km}) sin (2 3^{\circ}), 8 2^{\circ}

Simplify

\frac{250 km}{590 km} sin (2 3^{\circ}) : \frac{25 sin ( 2 3 ^{\circ} )}{59}

\frac{250 km}{590 km} sin (2 3^{\circ})

Multiply fractions:

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

= \frac{250 km sin ( 2 3 ^{\circ} )}{590 km}

Cancel the common factor:

k

= \frac{250 m sin ( 2 3 ^{\circ} )}{590 m}

Cancel the common factor:

m

= \frac{250 sin ( 2 3 ^{\circ} )}{590}

Cancel the common factor:

10

= \frac{25 sin ( 2 3 ^{\circ} )}{59}

Apply trig inverse properties

sin (b) = \frac{25 sin ( 2 3 ^{\circ} )}{59}

General solutions for

sin (b) = \frac{25 sin ( 2 3 ^{\circ} )}{59}

sin (x) = a \Rightarrow x = arcsin (a) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (a) + 36 0^{\circ} n

b = arcsin (\frac{25 sin ( 2 3 ^{\circ} )}{59}) + 36 0^{\circ} n, b = 18 0^{\circ} - arcsin (\frac{25 sin ( 2 3 ^{\circ} )}{59}) + 36 0^{\circ} n

b = arcsin (\frac{25 sin ( 2 3 ^{\circ} )}{59}) + 36 0^{\circ} n, b = 18 0^{\circ} - arcsin (\frac{25 sin ( 2 3 ^{\circ} )}{59}) + 36 0^{\circ} n

Solutions for the range

8 2^{\circ}

No Solution for b \in R

sin (x) = \frac{30}{60}

4 cos (z) + 5 = 0

y, sin (x^{2} y^{2}) = x

5 + 4 cos (θ) = 0

- sec^{2} (x) + tan (x) + 1 = 0

What is the general solution for sin(b)=((250km)/(590km))sin(23),82 ?
The general solution for sin(b)=((250km)/(590km))sin(23),82 is No Solution for b\in\mathbb{R}