What is the general solution for sin(2θ)= 12/13 ?

The general solution for sin(2θ)= 12/13 is θ=(1.17600…)/2+pin,θ= pi/2-(1.17600…)/2+pin

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sin(2θ)= 12/13

sin (2 θ) = \frac{12}{13}

θ = \frac{1.17600 \dots}{2} + πn, θ = \frac{π}{2} - \frac{1.17600 \dots}{2} + πn

Solution steps

sin (2 θ) = \frac{12}{13}

Apply trig inverse properties

2 θ = arcsin (\frac{12}{13}) + 2 πn, 2 θ = π - arcsin (\frac{12}{13}) + 2 πn

Solve

2 θ = arcsin (\frac{12}{13}) + 2 πn : θ = \frac{arcsin ( \frac{12}{13} )}{2} + πn

Solve

2 θ = π - arcsin (\frac{12}{13}) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( \frac{12}{13} )}{2} + πn

θ = \frac{arcsin ( \frac{12}{13} )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( \frac{12}{13} )}{2} + πn

Show solutions in decimal form

θ = \frac{1.17600 \dots}{2} + πn, θ = \frac{π}{2} - \frac{1.17600 \dots}{2} + πn

sin^{2} (θ) + sin (θ) = 2

7 sin^{2} (θ) - 3 sin (θ) = 4

- 5 - 4 tan (θ) = - 1

sin (θ) = - \frac{7}{10}

10 cos (2 x) - 3 = 2

What is the general solution for sin(2θ)= 12/13 ?
The general solution for sin(2θ)= 12/13 is θ=(1.17600…)/2+pin,θ= pi/2-(1.17600…)/2+pin