What is the general solution for solvefor x,1sin(25)=1.51sin(x) ?

The general solution for solvefor x,1sin(25)=1.51sin(x) is x=0.28366…+360n,x=180-0.28366…+360n

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solvefor x,1sin(25)=1.51sin(x)

solve for

x, 1 sin (2 5^{\circ}) = 1.51 sin (x)

x = 0.28366 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.28366 \dots + 36 0^{\circ} n

Radians

x = 0.28366 \dots + 2 πn, x = π - 0.28366 \dots + 2 πn

Solution steps

1 \cdot sin (2 5^{\circ}) = 1.51 sin (x)

Switch sides

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

Multiply:

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

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

Divide both sides by

1.51

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

Divide both sides by

1.51

\frac{1.51 sin ( x )}{1.51} = \frac{sin ( 2 5 ^{\circ} )}{1.51}

Simplify

sin (x) = \frac{sin ( 2 5 ^{\circ} )}{1.51}

sin (x) = \frac{sin ( 2 5 ^{\circ} )}{1.51}

Apply trig inverse properties

sin (x) = \frac{sin ( 2 5 ^{\circ} )}{1.51}

General solutions for

sin (x) = \frac{sin ( 2 5 ^{\circ} )}{1.51}

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} )}{1.51}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{sin ( 2 5 ^{\circ} )}{1.51}) + 36 0^{\circ} n

x = arcsin (\frac{sin ( 2 5 ^{\circ} )}{1.51}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{sin ( 2 5 ^{\circ} )}{1.51}) + 36 0^{\circ} n

Show solutions in decimal form

x = 0.28366 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.28366 \dots + 36 0^{\circ} n

What is the general solution for solvefor x,1sin(25)=1.51sin(x) ?
The general solution for solvefor x,1sin(25)=1.51sin(x) is x=0.28366…+360n,x=180-0.28366…+360n