What is the general solution for 1.31sin(35)=2.42sin(x) ?

The general solution for 1.31sin(35)=2.42sin(x) is x=0.31570…+360n,x=180-0.31570…+360n

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1.31sin(35)=2.42sin(x)

1.31 sin (3 5^{\circ}) = 2.42 sin (x)

x = 0.31570 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.31570 \dots + 36 0^{\circ} n

Radians

x = 0.31570 \dots + 2 πn, x = π - 0.31570 \dots + 2 πn

Solution steps

1.31 sin (3 5^{\circ}) = 2.42 sin (x)

Switch sides

2.42 sin (x) = 1.31 sin (3 5^{\circ})

Divide both sides by

2.42

2.42 sin (x) = 1.31 sin (3 5^{\circ})

Divide both sides by

2.42

\frac{2.42 sin ( x )}{2.42} = \frac{1.31 sin ( 3 5 ^{\circ} )}{2.42}

Simplify

sin (x) = \frac{1.31 sin ( 3 5 ^{\circ} )}{2.42}

sin (x) = \frac{1.31 sin ( 3 5 ^{\circ} )}{2.42}

Apply trig inverse properties

sin (x) = \frac{1.31 sin ( 3 5 ^{\circ} )}{2.42}

General solutions for

sin (x) = \frac{1.31 sin ( 3 5 ^{\circ} )}{2.42}

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

x = arcsin (\frac{1.31 sin ( 3 5 ^{\circ} )}{2.42}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{1.31 sin ( 3 5 ^{\circ} )}{2.42}) + 36 0^{\circ} n

x = arcsin (\frac{1.31 sin ( 3 5 ^{\circ} )}{2.42}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{1.31 sin ( 3 5 ^{\circ} )}{2.42}) + 36 0^{\circ} n

Show solutions in decimal form

x = 0.31570 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.31570 \dots + 36 0^{\circ} n

cos (x + 6 0^{\circ}) = \frac{1}{2}

3 cos (θ) = - cos (θ) tan (θ)

cos (2 x) = cos (x) - cos (3 0^{\circ}) [180]

1 - 2 tan (x) = tan^{2} (x)

- 13 a \cdot sin (2 x) - 13 b \cdot cos (2 x) = 0

What is the general solution for 1.31sin(35)=2.42sin(x) ?
The general solution for 1.31sin(35)=2.42sin(x) is x=0.31570…+360n,x=180-0.31570…+360n