What is the general solution for 1sin(40)=1.5sin(x) ?

The general solution for 1sin(40)=1.5sin(x) is x=0.44285…+360n,x=180-0.44285…+360n

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1sin(40)=1.5sin(x)

1 sin (4 0^{\circ}) = 1.5 sin (x)

x = 0.44285 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.44285 \dots + 36 0^{\circ} n

Radians

x = 0.44285 \dots + 2 πn, x = π - 0.44285 \dots + 2 πn

Solution steps

1 \cdot sin (4 0^{\circ}) = 1.5 sin (x)

Switch sides

1.5 sin (x) = 1 \cdot sin (4 0^{\circ})

Multiply:

1 \cdot sin (4 0^{\circ}) = sin (4 0^{\circ})

1.5 sin (x) = sin (4 0^{\circ})

Divide both sides by

1.5

1.5 sin (x) = sin (4 0^{\circ})

Divide both sides by

1.5

\frac{1.5 sin ( x )}{1.5} = \frac{sin ( 4 0 ^{\circ} )}{1.5}

Simplify

sin (x) = \frac{sin ( 4 0 ^{\circ} )}{1.5}

sin (x) = \frac{sin ( 4 0 ^{\circ} )}{1.5}

Apply trig inverse properties

sin (x) = \frac{sin ( 4 0 ^{\circ} )}{1.5}

General solutions for

sin (x) = \frac{sin ( 4 0 ^{\circ} )}{1.5}

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

x = arcsin (\frac{sin ( 4 0 ^{\circ} )}{1.5}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{sin ( 4 0 ^{\circ} )}{1.5}) + 36 0^{\circ} n

x = arcsin (\frac{sin ( 4 0 ^{\circ} )}{1.5}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{sin ( 4 0 ^{\circ} )}{1.5}) + 36 0^{\circ} n

Show solutions in decimal form

x = 0.44285 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.44285 \dots + 36 0^{\circ} n

4 sin^{2} (x) = 4 + cos (x)

3 sin (2 x) - 9 sin (x) - 4 cos (x) + 6 = 0

\frac{10}{sin ( x )} = \frac{13}{sin ( 7 2 ^{\circ} )}

sin (x) - 3 cos (x) = 1, 0 \leq x < 2 π

sin (α) + 1 = cos (α)

What is the general solution for 1sin(40)=1.5sin(x) ?
The general solution for 1sin(40)=1.5sin(x) is x=0.44285…+360n,x=180-0.44285…+360n