What is the general solution for sin(θ)=(12sin(43))/9 ?

The general solution for sin(θ)=(12sin(43))/9 is θ=1.14167…+360n,θ=180-1.14167…+360n

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sin(θ)=(12sin(43))/9

sin (θ) = \frac{12 sin ( 4 3 ^{\circ} )}{9}

θ = 1.14167 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 1.14167 \dots + 36 0^{\circ} n

Radians

θ = 1.14167 \dots + 2 πn, θ = π - 1.14167 \dots + 2 πn

Solution steps

sin (θ) = \frac{12 sin ( 4 3 ^{\circ} )}{9}

Cancel the common factor:

3

sin (θ) = \frac{4 sin ( 4 3 ^{\circ} )}{3}

Apply trig inverse properties

sin (θ) = \frac{4 sin ( 4 3 ^{\circ} )}{3}

General solutions for

sin (θ) = \frac{4 sin ( 4 3 ^{\circ} )}{3}

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

θ = arcsin (\frac{4 sin ( 4 3 ^{\circ} )}{3}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{4 sin ( 4 3 ^{\circ} )}{3}) + 36 0^{\circ} n

θ = arcsin (\frac{4 sin ( 4 3 ^{\circ} )}{3}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{4 sin ( 4 3 ^{\circ} )}{3}) + 36 0^{\circ} n

Show solutions in decimal form

θ = 1.14167 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 1.14167 \dots + 36 0^{\circ} n

2^{t a n (α)} = \frac{1}{4}

tan^{4} (x) - 3 tan^{2} (x) + 1 = 0

tan (θ) = \frac{- 4}{- 1}

cos (2 x) - 7 cos (x) + 6 = 0

csc^{2} (x) = 2 + csc (x)

What is the general solution for sin(θ)=(12sin(43))/9 ?
The general solution for sin(θ)=(12sin(43))/9 is θ=1.14167…+360n,θ=180-1.14167…+360n