What is the general solution for sin(x)=(48sin(69))/(47.5) ?

The general solution for sin(x)=(48sin(69))/(47.5) is x=1.23275…+360n,x=180-1.23275…+360n

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sin(x)=(48sin(69))/(47.5)

sin (x) = \frac{48 sin ( 6 9 ^{\circ} )}{47.5}

x = 1.23275 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 1.23275 \dots + 36 0^{\circ} n

Radians

x = 1.23275 \dots + 2 πn, x = π - 1.23275 \dots + 2 πn

Solution steps

sin (x) = \frac{48 sin ( 6 9 ^{\circ} )}{47.5}

Apply trig inverse properties

sin (x) = \frac{48 sin ( 6 9 ^{\circ} )}{47.5}

General solutions for

sin (x) = \frac{48 sin ( 6 9 ^{\circ} )}{47.5}

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

x = arcsin (\frac{48 sin ( 6 9 ^{\circ} )}{47.5}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{48 sin ( 6 9 ^{\circ} )}{47.5}) + 36 0^{\circ} n

x = arcsin (\frac{48 sin ( 6 9 ^{\circ} )}{47.5}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{48 sin ( 6 9 ^{\circ} )}{47.5}) + 36 0^{\circ} n

Show solutions in decimal form

x = 1.23275 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 1.23275 \dots + 36 0^{\circ} n

2 cos^{2} (3 x) + cos (3 x) - 1 = 0

sin (3 x) = \frac{- 3}{4}

4 cos (x - 4 5^{\circ}) + 7 = 10

cos^{2} (x) = 3 (1 - sin (x))

- 2 = 2 cos (3 θ)

What is the general solution for sin(x)=(48sin(69))/(47.5) ?
The general solution for sin(x)=(48sin(69))/(47.5) is x=1.23275…+360n,x=180-1.23275…+360n