What is the general solution for 1sin(25)=1.506sin(x) ?

The general solution for 1sin(25)=1.506sin(x) is x=0.28444…+360n,x=180-0.28444…+360n

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1sin(25)=1.506sin(x)

1 sin (2 5^{\circ}) = 1.506 sin (x)

x = 0.28444 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.28444 \dots + 36 0^{\circ} n

Radians

x = 0.28444 \dots + 2 πn, x = π - 0.28444 \dots + 2 πn

Solution steps

1 \cdot sin (2 5^{\circ}) = 1.506 sin (x)

Switch sides

1.506 sin (x) = 1 \cdot sin (2 5^{\circ})

Multiply:

1 \cdot sin (2 5^{\circ}) = sin (2 5^{\circ})

1.506 sin (x) = sin (2 5^{\circ})

Divide both sides by

1.506

1.506 sin (x) = sin (2 5^{\circ})

Divide both sides by

1.506

\frac{1.506 sin ( x )}{1.506} = \frac{sin ( 2 5 ^{\circ} )}{1.506}

Simplify

sin (x) = \frac{sin ( 2 5 ^{\circ} )}{1.506}

sin (x) = \frac{sin ( 2 5 ^{\circ} )}{1.506}

Apply trig inverse properties

sin (x) = \frac{sin ( 2 5 ^{\circ} )}{1.506}

General solutions for

sin (x) = \frac{sin ( 2 5 ^{\circ} )}{1.506}

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 ( 2 5 ^{\circ} )}{1.506}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{sin ( 2 5 ^{\circ} )}{1.506}) + 36 0^{\circ} n

x = arcsin (\frac{sin ( 2 5 ^{\circ} )}{1.506}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{sin ( 2 5 ^{\circ} )}{1.506}) + 36 0^{\circ} n

Show solutions in decimal form

x = 0.28444 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.28444 \dots + 36 0^{\circ} n

\frac{1}{2} tan (2 θ) = tan (θ)

csc (x) - 3 = - 2 sin (x)

\frac{sin ( y )}{80} = \frac{sin ( 4 0 ^{\circ} )}{60}

cos (w) = 0.75 sin (9 0^{\circ} - w)

csc (4 x) = 2

What is the general solution for 1sin(25)=1.506sin(x) ?
The general solution for 1sin(25)=1.506sin(x) is x=0.28444…+360n,x=180-0.28444…+360n